lecture notes in linear algebralecture notes in linear algebra dr. abdullah al-azemi mathematics...

Lecture Notes in Linear Algebra

Dr. Abdullah Al-Azemi

Mathematics Department

Kuwait University

February 4, 2017

Contents

1 Linear Equations and Matrices 1

1.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 Special types of matrices: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.2 Matrix Operation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Dot Product and Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4.1 Properties of matrix-scalar multiplication . . . . . . . . . . . . . . . . . . . . . . 13

1.4.2 Properties of Matrix Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4.3 Properties of Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4.4 Properties of Matrix Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4.5 Symmetric and Skew-Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . 16

1.6 Solutions of Linear System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.6.1 Solving System of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.7 The Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2 Determinants 43

2.1 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.2 Cofactor Expansion and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3 Vectors in Rn 65

3.2 Vectors in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

i

ii CONTENTS

3.2.1 Vectors Operations: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.2.2 The Dot Product of Vectors: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.2.3 Properties of Dot Product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.2.4 The Angle Between Two Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.2.5 Principle of Unit Vectors in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5 Real Vector Spaces 77

5.1 Real Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

5.2 Real Vector Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.3 Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

5.4 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

5.5 Homogenous System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.6 The Rank of a Matrix and its Application . . . . . . . . . . . . . . . . . . . . . . . . . 103

6 Diagonalization 111

6.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

6.2 Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.2.1 Case 1: Diagonalization of a Matrix with Distinct Eigenvalues . . . . . . . . . . 118

6.2.2 Case 2: Diagonalization of a Matrix with Repeated Eigenvalues . . . . . . . . . 119

4 Vectors in Rn 125

4.1 The Cross Product in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

4.2 Planes and Lines in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

1

Chapter

Linear Equations and Matrices

1.2 Matrices

Definition 1.2.1

An m× n matrix A is a rectangular array of m · n real numbers arranged in m horizontal rows

and n vertical columns. That is,

A =

a11 a12 . . . a1j . . . a1n

a21 a22 . . . a2j . . . a2n...

......

......

...

ai1 ai2 . . . aij . . . ain...

......

......

...

am1 am2 . . . amj . . . amn

In the matrix A above, we have:

• a11, a12, . . . , amn are called elements (or entries) of the matrix.

• The entry aij lies in the intersection of row i and column j.

• The size (or order) of A is m by n, written as m× n.

• The matrix is called a square matrix if m = n.

• We write Mm×n for the class of all real matrices of size m× n.

If a matrix A is of size 1×n, then we say that A is a vector of size (length) n and we write A ∈ Rn.

In addition, if A is of size n× 1, we say that A ∈ Rn.

Example 1.2.1

A matrix A ∈M2×3 and another matrix B ∈M3×3.

A =

2 3 −2

4 1 −1

2×3

B =

3 0 1

4 2 0

−1 4 1

3×3

1

2 Chapter 1. Linear Equations and Matrices

An m× n matrix can be represented by the following ways:

• A = (aij) for 1 ≤ i ≤ m and 1 ≤ j ≤ n,

• A =

row 1

row 2...

row m

, where row i = [ai1 ai2 . . . ain], for 1 ≤ i ≤ m, or

• A = [col 1 col 2 . . . col n], where col j =

a1j

a2j...

amj

for 1 ≤ j ≤ n.

Example 1.2.2

List all of the entries of the matrix A = (aij) ∈ R3, for 1 ≤ i ≤ 3 and 1 ≤ j ≤ 3, where

aij =

i + j, if i > j

0, if i = j

i− j, if i < j

Solution:

We list the entries of A in the following way:

A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

0 −1 −2

3 0 −1

4 5 0

1.2.1 Special types of matrices:

1. The ”zero matrix”, say O, is an m× n matrix whose entries are all equal to 0.

O =

0 · · · 0...

. . ....

0 · · · 0

1.2. Matrices 3

2. The ”diagonal matrix”, D, is a square matrix in Rn, where D = (dij) so that dij = 0 for all

i 6= j.

D =

d11 0. . .

0 dnn

3. The ”scalar matrix” is a diagonal matrix S whose all diagonal entries are equal to some a scalar

c.

S =

c 0. . .

0 c

4. The ”identity matrix” (or unit matrix) is a scalar matrix In whose all diagonal entries are

equal to 1.

In =

1 0. . .

0 1

5. The ”lower triangular matrix”, L, is a square matrix in Rn, where L = (lij) so that lij = 0 for

all i < j.

L =

l11 0...

. . .

ln1 . . . lnn

6. The ”upper triangular matrix”, U , is a square matrix in Rn, where U = (uij) so that uij = 0

for all i > j.

U =

u11 . . . u1n. . .

...

0 unn

Remark 1.2.1. Two m × n matrices A = (aij) and B = (bij) are said to be equal if aij = bij for all

1 ≤ i ≤ m and 1 ≤ j ≤ n.

Example 1.2.3

Find the values of a, b, c, and d if

a + b c + d

c− d a− b

=

1 2

−2 1

.

Solution:


Since the two matrices are equal. Then

a + b = 1

a− b = 1

+ =⇒ 2a = 2 =⇒ a = 1 and hence b = 0.

c + d = 2

c− d = −2

+ =⇒ 2c = 0 =⇒ c = 0 and hence d = 2.

1.2.2 Matrix Operation:

• Matrix transpose

Definition 1.2.2

If A = (aij) is an m × n matrix, then the n × m matrix AT = (aji), where 1 ≤ j ≤ n and

1 ≤ i ≤ m, is called the transpose of A.

Example 1.2.1. If A =

a b c

d e f

2×3

, then AT =

a d

b e

c f

3×2

.

Example 1.2.4

(Q#6

pg. 20

)

If A is an n× n lower triangular matrix, then AT is an n× n upper triangular matrix.

Solution:

Since A is a lower triangular, then aij = 0 for all i < j. Therefore, the transpose of A, AT = (aji)

has entries aji = 0 if j > i and thus it is an upper triangular.

• Matrix addition and substraction

To add two m × n matrices A = (aij) and B = (bij), we must have the size(A) = size(B). Then,

A±B = C, where C = (cij) with cij = aij ± bij, for all 1 ≤ i ≤ m and 1 ≤ j ≤ n.

1.2. Matrices 5

Example 1.2.5

If possible, find A + B and AT −B, where

A =

2 1

3 0

−1 1

3×2

, and B =

1 2 4

5 1 1

2×3

.

Solution:

Clearly, A+B is not possible as they have different sizes. On the other hand, size(AT ) = size(B),

and

AT −B =

2 3 −1

1 0 1

2×3

−

1 2 4

5 1 1

2×3

=

1 1 −5

−4 −1 0

2×3

.

Example 1.2.6

(Q#5pg.20

)

Show that the sum and the difference of two lower triangular matrices is a lower triangular

matrix.

Solution:

Let A = (aij) and B = (bij) be two n × n matrices so that both are lower triangular matrices.

Thus, aij = 0 and bij = 0 for all i < j. If (cij) = C = A±B with cij = aij ± bij, then cij = 0 for

all i < j. Therefore, C is again a lower triangular matrix.

Remark 1.2.1

The sum and the difference of two upper triangular matrices is an upper triangular matrix.

• Scalar multiplication

If A = (aij) is an m × n matrix and c is a real number, then c A = (c · aij) for all 1 ≤ i ≤ m and

1 ≤ j ≤ n.


1.3 Dot Product and Matrix Multiplication

Definition 1.3.1

The dot (inner) product of the n-vectors x = (x1 x2 · · · xn) and y = (y1 y2 · · · yn) is

defined by

x · y = x1y1 + x2y2 + · · ·+ xnyn =n∑

i=1

xiyi. (1.3.1)

An n-vector x = (x1 x2 · · · xn) can be written as x = (x1, x2, · · · , xn).

Example 1.3.1

Find x · y if x = (1, 0, 2,−1) and y = (3, 5,−1, 4).

Solution:

Clearly, x · y = 3 + 0 + (−2) + (−4) = −3.

Definition 1.3.2

Let A = (aij) be an m× p matrix and B = (bij) be a p× n matrix. Then, the product of A and

B, (AB), is the m× n matrix C = (cij) = rowi(A) · colj(B) where

cij = ai1b1j + ai2b2j + · · ·+ aipbpj =p

∑

k=1

aikbkj, for 1 ≤ i ≤ m and 1 ≤ j ≤ n.

Note that the entry (i, j) of the product AB equals to the dot product of rowi(A) by the colj(B).

Remark 1.3.1

To get the product AB, we must have ”# of columns in A = # of rows in B”.

1.3. Dot Product and Matrix Multiplication 7

Example 1.3.2

If A =

a11 a12 a13

a21 a22 a23

2×3

, and B =

b11 b12

b21 b22

b31 b32

3×2

, then

AB =

a11b11 + a12b21 + a13b31 a11b12 + a12b22 + a13b32

a21b11 + a22b21 + a23b31 a21b12 + a22b22 + a23b32

2×2

Example 1.3.3

Find AB, ABT , and AT B if possible, where

A =

1 0 2

1 −2 0

, and B =

4 1 −1

2 5 1

.

Solution:

• Clearly, AB is not possible to compute since # columns in A is not the same as # rows of

B.

• ABT : excercise.

• AT B =

1 1

0 −2

2 0

4 1 −1

2 5 1

=

4 + 2 1 + 5 −1 + 1

−4 −10 −2

8 2 −2

=

6 6 0

−4 −10 −2

8 2 −2

Remark 1.3.2

In general, AB 6= BA. For instance, consider A =

1 0

1 0

and B =

0 0

1 1

. Check it yourself!!

It is always true that AB = BA if A = In or that A = B.


Example 1.3.4

Let A =[

1 −2 0

]

. Find all values of c so that c A · AT = 15.

Solution:

Note that cA · AT = c[

1 −2 0

]

·

1

−2

0

= 5c. Therefore, 5c = 15 and hence c = 3.

Example 1.3.5

Let A =

1 −2 3

4 2 1

0 1 −2

, and B =

1 4

3 −1

−2 2

. Compute the (3, 2)-entry of AB.

Solution:

The (3, 2)-entry = row3(A) · col2(B) = [0 1 − 2] · [4 − 1 2] = −1− 4 = −5.

Definition 1.3.3

Let A be an m× n matrix and x ∈ Rn (be an n-vector), then

Ax =

a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...

am1 am2 · · · amn

x1

x2...

xn

=

a11x1 + a12x2 + · · ·+ a1nxn

a21x1 + a22x2 + · · ·+ a2nxn...

am1x1 + am2x2 + · · ·+ amnxn

=

row1(A) · xrow2(A) · x

...

rowm(A) · x

= x1

a11

a21...

am1

+ x2

a12

a22...

am2

+ · · ·+ xn

a1n

a2n...

amn

= x1col1(A) + x2col2(A) + · · ·+ xncoln(A).

That is, Ax is a linear combination of columns of A and the entries of x are the coefficients.


Example 1.3.6

Write the following product as a linear combination of the columns of the first matrix.

1 3

2 1

4 2

3

−2

= 3

1

2

4

+ (−2)

3

1

2

.

Example 1.3.7

(Q#9pg.38

)

If A is an m× p matrix and B is a p× n matrix, then

(a) Show that colj(AB) = A colj(B).

(b) Show that rowi(AB) =(

rowi(A))

B.

Solution:

(a) Let C = AB. Clearly for 1 ≤ j ≤ n,

colj(AB) = colj(C) =

c1j

c2j...

cmj

=

row1(A) · colj(B)

row2(A) · colj(B)...

rowm(A) · colj(B)

=

row1(A)

row2(A)...

rowm(A)

· colj(B) = A · colj(B).

(b) Let C = AB. Clearly for 1 ≤ i ≤ m,

rowi(AB) = rowi(C) =[

ci1 ci2 · · · cin

]

=[

rowi(A) · col1(B) rowi(A) · col2(B) · · · rowi(A) · coln(B)

]

= rowi(A) ·[

col1(B) col2(B) · · · coln(B)

]

=(

rowi(A))

·B.

TRUE or FALSE:

⋆ If a matrix B has a column of zeros, then the product AB has a column of zeros as well. (TRUE).

reason: Assume that column j of B is a column of zero. Then, colj(AB) = A colj(B) = 0.


Example 1.3.8

Let A =

1 2

3 4

−1 5

3×2

and B =

−2 3 4

3 2 1

2×3

. Find the columns of AB as a linear combination

of columns of A.

Solution:

Clearly,

• col1(AB) = A col1(B) =

1 2

3 4

−1 5

−2

3

= −2

1

3

−1

+ 3

2

4

5


1 2

3 4

−1 5

3

2

= 3

1

3

−1

+ 2

2

4

5


1 2

3 4

−1 5

4

1

= 4

1

3

−1

+ 1

2

4

5

Example 1.3.9

Compute A2017, where A =

1 1

0 1

.

Solution:

Here, we will compute some powers of the matrix A to look for a pattern. That is,

A2 = AA =

1 1

0 1

1 1

0 1

=

1 2

0 1

, A3 = A2A =

1 2

0 1

1 1

0 1

=

1 3

0 1

A4 = A3A =

1 3

0 1

1 1

0 1

=

1 4

0 1

, · · ·An = An−1A =

1 (n− 1)

0 1

1 1

0 1

=

1 n

0 1

.

Therefore, A2017 = A2016 A =

1 2017

0 1

.


Matrix Form of Linear System of Equations:

Given a linear system of equation on n variables as in (1.3.2)

a11x1 + a12x2 + a13x3 + · · · + a1nxn = b1

a21x1 + a22x2 + a23x3 + · · · + a2nxn = b2...

am1x1 + am2x2 + am3x3 + · · · + amnxn = bm

(1.3.2)

we transform this system into the matrix form as in (1.3.3)

a11 a12 · · · a1n

a21 a22 · · · a2n...

......

...


x1

x2...

xn

=

b1

b2...

bm

(1.3.3)

We say that the form of (1.3.3) is the matrix form A x = B, where x is called the vector of

unknowns. Moreover, we define the augmented form as

[

A |B]

. (1.3.4)

Example 1.3.10

Write the following system of linear equations in the matrix form and in the augmented form.

2x − z + 3w = 1

x + 5y = −1

x + y + z = 0

Solution:

The matrix form is:

2 0 −1 3

1 5 0 0

1 1 1 0

x

y

z

w

=

1

−1

0

while the augmented form is given by

2 0 −1 3 1

1 5 0 0 −1

1 1 1 0 0

.


Exercise 1.3.1

1. Let A =

1 2

0 −1

. Compute A2 + I2.

2.(

Q#6pg.31

)

: The product of two upper triangular matrices is an upper triangular matrix.

Solution: The proof is NOT required.

3. Let A =

1 2

4 5

3 6

and B =

0 1 −3

−3 1 4

. Express the third row of AB as a linear

combination of the rows of B.

4. Let A =

1 1

0 2

2 0

and B =

4 1 −1

2 5 1

. Find AB and express the second column of AB

as a linear combination of the columns of A.

1.4. Properties of Matrix Operations 13

1.4 Properties of Matrix Operations

In what follows, let A, B, and C be matrices of appropriate sizes, and let r, s ∈ R.

1.4.1 Properties of matrix-scalar multiplication

1. r (s A) = (rs) A,

2. (r + s) A = r A + s A,

3. r (A + B) = r A + r B,

4. A(r B) = (r A)B = r (AB).

Example 1.4.1

(Q#32pg.51

)

If the m × n linear system A x = B has more than one solution, then it has infinitely

many solutions.

Solution:

Let y and z be two solutions to the system A x = B. Then, for r, s ∈ R, we have

A y = B

r A y = r B

A(r y) = r B. (1.4.1)

Similarly, we have

A(s z) = s B. (1.4.2)

Adding up (1.4.1) and (1.4.2), we got

A(r y + s z) = (r + s)B.

Therefore, (r y + s z) is a solution to the system whenever, r + s = 1. The result follows.


TRUE or FALSE:

⋆ If x1 and x2 are two solutions for Ax = B, then 32x1 − 2x2 is also a solution. (FALSE).

reason: Since 32− 2 = −1

26= 1.

1.4.2 Properties of Matrix Addition

1. A + B = B + A,

2. (A + B) + C = A + (B + C),

3. there is a unique matrix O such that O + A = A + O = A, the matrix O is called the

additive identity,

4. there is a unique matrix D such that A + D = D + A = O. In particular, D = −A is called

the additive inverse of A.

1.4.3 Properties of Matrix Multiplication

1. A(BC) = (AB)C,

2. A(B + C) = AB + AC,

3. (A + B)C = AC + BC,

4. there is a unique (identity) matrix I such that AI = IA = A.

Remark 1.4.1

Let p, q, and r be rational numbers and let A, B ∈Mm×n, then

1. ApAq = Ap+q,

2. (Ap)q = Apq,

3. AB 6= BA,


4. (AB)p 6= ApBp,

5. for a, b ∈ R: ab = 0 only if a = 0 or b = 0. This is not true ingeneral for matrices. For

instance, if A = [ 0 10 1 ] , and B = [ 1 1

0 0 ], then AB = O where A and B are both non-zero

matrices.

1.4.4 Properties of Matrix Transpose

1. (AT )T = A,

2. (A + B)T = AT + BT ,

3. (AB)T = BT AT ,

4. for any real number r ∈ R, (rA)T = rAT ,

Note That: Property 3 can be generalized to (ABC · · ·XY Z)T = ZT Y T XT · · ·CT BT AT .

Example 1.4.2

Let A =

2 0

1 −1

and B =

1 4

3 5

. Find AT , BT , AB, (AB)T , and BT AT .

Solution:

Clearly, AT =

2 1

0 −1

and BT =

1 3

4 5

.

AB =

2 + 0 8 + 0

1− 3 4− 5

=

2 8

−2 −1

⇒ (AB)T =

2 −2

8 −1

Moreover,

BT AT =

2 + 0 1− 3

8 + 0 4− 5

=

2 −2

8 −1


Example 1.4.3

(T #9pg.51

)

Find a 2× 2 matrix B 6= O and B 6= I2 so that AB = BA if A =

1 2

0 1

Solution:

Let B =

a b

c d

. Therefore,

AB = BA ⇒

a + 2c b + 2d

c d

=

a 2a + b

c 2c + d

,

thus a = a + 2c→ c = 0 and b + 2d = 2a + b→ a = d. Let s, t ∈ R so that a = s and b = t to

get B =

s t

0 s

. Note that s and t can not be both 0 since B 6= O, and the case of t = 0 and

s = 1 can NOT be hold since B 6= I.

1.4.5 Symmetric and Skew-Symmetric Matrices

Definition 1.4.1

Let A ∈ Mn×n. A is called symmetric matrix whenever AT = A, and it is called skew-

symmetric matrix whenever AT = −A.

For instance, A =

1 2 3

2 4 5

3 5 0

is a symmetric matrix; where B =

0 2

−2 0

is a skew-symmetric

matrix.

Example 1.4.4

(T #23pg.51

)

Let A and B be two symetric matrices. Show that

(a) A + B is symmetric.

(b) AB is symmetric (↔) if and only if AB = BA.


Solution:

Since A and B are both symmetirc, then AT = A and BT = B.

(a): Clearly, (A + B)T = AT + BT = A + B. Therefore, A + B is symmetric.

(b): We first prove (→). Let AB be symmetirc. Thus AB = (AB)T = BT AT = BA which

shows that AB = BA. To show the other direction (←), assume that AB = BA. Thus,

(AB)T = BT AT = BA = AB which shows that AB is symmetric.

Example 1.4.5

If A and B are skew-symmetric matrices and AB = BA, then AB is symmetric.

Solution:

It is given that AT = −A and BT = −B; thus, (AB)T = BT AT = (−B)(−A) = BA = AB.

Therefore, AB is symmetric.

Example 1.4.6

(T #26pg.51

)

Show that AT A and AAT are both symmetric matrices.

Solution:

It is clear that (AT A)T = AT (AT )T = AT A which shows that AT A is symmetric. In addition,

(AAT )T = (AT )T AT = AAT shows that AAT is also symmetric.

Example 1.4.7

(T #27pg.51

)

Let A ∈Mn×n. Show that

(a) AT + A is symmetirc.

(b) A− AT is skew-symmetric.

Solution:

[(a)]: (AT + A)T = (AT )T + AT = A + AT = AT + A. Therefore, AT + A is symmetric.

[(b)]: (A−AT )T = AT − (AT )T = AT −A = −(A−AT ). Therefore, A−AT is skew-symmetric.


Example 1.4.8

(T #28pg.51

)

Let A ∈Mn×n. Then A can be written as A = S + K, where S is symmetric matrix and

K is skew-symmetric matrix.

Solution:

From Example 1.4.7, we know that AT + A is symmetric and that A − AT is skew-symmetric.

If S = 12(AT + A) which is symmetric and K = 1

2(A− AT ) which is skew-symmetric, then

S + K =1

2(AT + A) +

1

2(A− AT ) =

1

2AT +

1

2A +

1

2A− 1

2AT = A.


Exercise 1.4.1

1. Let A =

1 1

0 1

.

(a) Find A1977.

(b) Find all matrices B such that AB = BA.

2. Let A =

12

12

12

12

. Find A100.

3. Show that if A and B are symmetric matrices, then AB−BA is a skew-symmetric matrix.

4. Let A be 2× 2 skew-symmetric matrix. If A2 = A, then A = 0.

5. Let A =

1

3

−1

3

. Find all constants c ∈ R such that (cA)T · (cA) = 5.

6. Let A =

2 −1 3

0 4 1

1 −2 −3

. Find a symmetric matrix S and a skew symmetric matrix K

such that A = S + K.


1.6 Solutions of Linear System of Equations

Definition 1.6.1

A matrix A ∈ Mm×n is said to be in the reduced row echelon form (r.r.e.f. for short) if it

satisfies the following conditions:

1. the row of zeros (if any) should be at the bottom of A,

2. the leading (first) entry of a non-zero row should be 1,

3. the leading entry of row i + 1 should be on the right of the leading entry of row i,

4. the columns that contain a leading entry 1, all its other entries are zeros.

The matrix A is said to be in the row echelon form (r.e.f.) if it does not satisfy the last Condition.

Example 1.6.1

Here are some examples of some matrices to clarify the previous conditions:

(1) A =

1 0 0 3 0

0 0 1 −3 0

0 0 0 0 0

is r.r.e.f.

(2) B =

1 3 0 4

0 0 0 0

0 0 1 −2

is not r.r.e.f. since it has a row of zeros in the second row.

(3) C =

1 0 4 5

0 2 −2 3

0 0 1 2

is not r.r.e.f. since the leading entry in the second row is not 1.

(4) D =

1 0 0 2

0 0 1 3

0 1 0 0

is not r.r.e.f. since the leading entry of the third row is on the left

of the leading entry of the second row. Switch 2nd and 3rd rows to get the r.r.e.f. form.

1.6. Solutions of Linear System of Equations 21

⋆ Given a matrix A ∈Mm×n, we define the following elementary row operations:

1. interchanging a row by another row,

2. multiplying a row by a non-zero scalar,

3. adding a multiple of a row to another row. That is,

replace ai1, ai2, · · · , ain by ai1 + cak1, ai2 + cak2, · · · , ain + cakn, for 1 ≤ i, k ≤ n.

Definition 1.6.2

An m × n matrix A is called row equivalent to an m × n matrix B if B can be obtained by

applying a finite sequence of elementary row operation to A. In this case, we write A ≈ B.

⋆ Properties of matrix equivalence: For any m× n matrices A, B, and C we have

1. A ≈ A,

2. A ≈ B ⇒ B ≈ A,

3. A ≈ B and B ≈ C ⇒ A ≈ C.

Theorem 1.6.1

Every non-zero m× n matrix is row equivalent to a unique matrix in the r.r.e.f.

Example 1.6.2

Find the matrix B which is in r.r.e.f. and is row equivalent to A =

1 2 3 9

2 −1 1 8

3 0 −1 3

.

Solution:

Ar2−2r1→r2−−−−−−→r3−3r1→r3

1 2 3 9

0 −5 −5 −10

0 −6 −10 −24

− 15

r2→r2−−−−−→− 1

2r3→r3

1 2 3 9

0 1 1 2

0 3 5 12

r1−2r2→r1−−−−−−→r3−3r2→r3

1 0 1 5

0 1 1 2

0 0 2 6


12

r3→r3−−−−→

1 0 1 5

0 1 1 2

0 0 1 3

r1−r3→r1−−−−−−→r2−r3→r2

1 0 0 2

0 1 0 −1

0 0 1 3

= B (in r.r.e.f.) ≈ A.

Example 1.6.3

Find the matrix B which is in r.r.e.f. and is row equivalent to A =

2 4 6 0

1 2 4 0

1 3 3 1

Solution:

A12

r1→r1−−−−→

1 2 3 0

1 2 4 0

1 3 3 1

r2−r1→r2−−−−−−→r3−r1→r3

1 2 3 0

0 0 1 0

0 1 0 1

r2↔r3−−−→

1 2 3 0

0 1 0 1

0 0 1 0

r1−2r2→r1−−−−−−→

1 0 3 −2

0 1 0 1

0 0 1 0

r1−3r3→r1−−−−−−→

1 0 0 −2

0 1 0 1

0 0 1 0

= B (in r.r.e.f.) ≈ A.

1.6.1 Solving System of Linear Equations

Solving Ax = B

Ax = O (homogenous)

non-trivial soltrivial sol

Ax = B where B 6= O (non-homogenous)

infinite solno solunique sol

Remark 1.6.1

⋆ Guass-Jordan method for solving Ax = B:

1. Find the r.r.e.f. of augmented matrix[

A |B]

,

2. Solve the reduced system.

Note that, the solution of the reduced system is the solution of the original one.


Theorem 1.6.2

Let Ax = B and Cx = D be two linear systems of equations. If [ A |B ] ≈ [ C |D ], then the two

system have the same solutions.

(α) Solving Non-Homogenous System Ax = B with B 6= O

Example 1.6.4

Solve the following system using the Guass-Jordan method.

x + 2y + 3z = 9

2x − y + z = 8

3x − z = 3

Solution:

1 2 3 9

2 −1 1 8

3 0 −1 3

r2−2r1→r2−−−−−−→r3−3r1→r3

1 2 3 9

0 −5 −5 −10

0 −6 −10 −24

− 15

r2→r2−−−−−→− 1

2r3→r3

1 2 3 9

0 1 1 2

0 3 5 12

r1−2r2→r1−−−−−−→r3−3r2→r3

1 0 1 5

0 1 1 2

0 0 2 6

12

r3→r3−−−−→

1 0 1 5

0 1 1 2

0 0 1 3

r1−r3→r1−−−−−−→r2−r3→r2

1 0 0 2

0 1 0 −1

0 0 1 3

Therefore, the reduced system is: x = 2, y = −1, and z = 3. Thus, x =

2

−1

3

is a unique

solution to the system.

Remark 1.6.2

The system Ax = B has a unique solution if and only if A ≈ I.


Example 1.6.5


x1 + x2 − x3 + 4x4 = 1

+ x2 − 3x3 + 4x4 = 0

2x1 + 2x2 − 2x3 + 8x4 = 2

Solution:

1 1 −1 4 1

0 1 −3 4 0

2 2 −2 8 2

r3−2r1→r3−−−−−−→

1 1 −1 4 1

0 1 −3 4 0

0 0 0 0 0

r1−r2→r1−−−−−−→

1 0 2 0 1

0 1 −3 4 0

0 0 0 0 0

Therefor, the reduced system is:

x1 + 2x3 = 1

x2 − 3x3 + 4x4 = 0⇒ x1 = 1− 2x3

x2 = 3x3 − 4x4

We now fix x3 = r and x4 = t for r, t ∈ R to get the following infinite many solutions:

x1 = 1− 2r

x2 = 3r − 4t⇒ x =

1− 2r

3r − 4t

r

t

for all r, t ∈ R.

Remark 1.6.3

The system Ax = B has infinity many solutions if the number of unknowns is more than the

number of equations.

Example 1.6.6


x1 + x2 − x3 + 4x4 = 1

+ x2 − 3x3 + 4x4 = 0

2x1 + 2x2 − 2x3 + 8x4 = 3

Solution:


1 1 −1 4 1

0 1 −3 4 0

2 2 −2 8 2

r3−2r1→r3−−−−−−→

1 1 −1 4 1

0 1 −3 4 0

0 0 0 0 1

At this point, it can be seen that the third row suggests that 0 = 1 which is not possible.

Therefore, this system has no solution.

Remark 1.6.4

The system Ax = B has no solution whenever A ≈ a matrix with a row of zeros while [ A |B ] ≈a matrix with no rows of zeros.

Example 1.6.7

Consider the system:

x + y + z = 2

x + 2y − z = 2

x + 2y + (a2 − 5)z = a

Find all values of a so that the system has:

(a) a unique solution.

(b) infinite many solutions.

(c) no solution.

Solution:

In this kind of questions, it is not necessary to get the r.r.e.f. So, we will try to focus on the last

row which contains the term ”a” as follows:

1 1 1 2

1 2 −1 2

1 2 a2 − 5 a

r2−r1→r2−−−−−−→r3−r1→r3

1 1 1 2

0 1 −2 0

0 1 a2 − 6 a− 2

r3−r2→r3−−−−−−→

1 1 1 2

0 1 −2 0

0 0 a2 − 4 a− 2

At this point, we have:

(a) a unique solution if a2 − 4 6= 0⇐⇒ a 6= ±2⇐⇒ a ∈ R\{−2, +2}.(b) infinite solutions if

{

a2−4 = 0 and a−2 = 0}

⇐⇒{

a = ±2 and a = +2}

⇐⇒ a = +2.

(c) no solution if{

a2 − 4 = 0 and a− 2 6= 0}

⇐⇒{

a = ±2 and a 6= +2}

⇐⇒ a = −2.


(β) Solving Homogenous System Ax = O

Example 1.6.8


x + 2y + 3z = 0

x + 3y + 2z = 0

2x + y − 2z = 0

Solution:

1 2 3 0

1 3 2 0

2 1 −2 0

r2−r1→r2−−−−−−→r3−2r1→r3

1 2 3 0

0 1 −1 0

0 −3 −8 0

r1−2r2→r1−−−−−−→r3+3r2→r3

1 0 5 0

0 1 −1 0

0 0 −11 0

−111

r3→r3−−−−−→

1 0 5 0

0 1 −1 0

0 0 1 0

r1−5r3→r1−−−−−−→r2+r3→r2

1 0 0 0

0 1 0 0

0 0 1 0

Therefore, the reduced system is: x = 0, y = 0, and z = 0. Thus, x =

0

0

0

is a trivial solution.

Remark 1.6.5

The system Ax = O has a trivial solution whenever A ≈ I.

Example 1.6.9


x1 + x2 − x3 + 4x4 = 0

+ x2 − 3x3 + 4x4 = 0

2x1 + 2x2 − 2x3 + 8x4 = 0

Solution:


1 1 −1 4 0

0 1 −3 4 0

2 2 −2 8 0

r3−2r1→r3−−−−−−→

1 1 −1 4 0

0 1 −3 4 0

0 0 0 0 0

r1−r2→r1−−−−−−→

1 0 2 0 0

0 1 −3 4 0

0 0 0 0 0

Therefor, the reduced system is:

x1 + 2x3 = 0

x2 − 3x3 + 4x4 = 0⇒ x1 = −2x3

x2 = 3x3 − 4x4

We now fix x3 = r and x4 = t for r, t ∈ R to get the following non-trivial solution:

x1 = −2r

x2 = 3r − 4t⇒ x =

−2r

3r − 4t

r

t

for all r, t ∈ R.

Remark 1.6.6

The system Ax = O has a non-trivial solution if the number of unknowns is greater than the

number of equations in the reduced system.

Definition 1.6.3

Linear systems with at least one solution are called consistent, and linear system with no

solutions are called inconsistent.

Example 1.6.10

Discuss the consistency of the following homogenous system:

x + y − z = 0

x − y + 3z = 0

x + y + (a2 − 5)z = 0

Solution:

1 1 −1 0

1 −1 3 0

1 1 a2 − 5 0

r2−r1→r2−−−−−−→r3−r1→r3

1 1 −1 0

0 −2 4 0

0 0 a2 − 4 0

.

The system has:


(a) trivial solution when a2 − 4 6= 0⇐⇒ a 6= ±2⇐⇒ a ∈ R\{±2},(b) non-trivial solution when a2 − 4 = 0⇐⇒ a = ±2.

Example 1.6.11

Discuss the consistency of the following non-homogenous system:

x + 3y + kz = 4

2x + ky + 12z = 6

Solution:

1 3 k 4

2 k 12 6

r2−2r1→r2−−−−−−→

1 3 k 4

0 k − 6 12− 2k −2

.

The system:

(a) can not have a unique solution (not equivalent to I because it is not square matrix!!),

(b) has infinite solutions when k 6= 6,

(c) has no solutions when k = 6.

Example 1.6.12

(T #11pg.89

)

Let u and v be two solutions of the homogenous system Ax = O. Show that ru + sv

(for r, s ∈ R) is a solution to the same system.

Solution:

Since u and v are two solution of Ax = O, then Au = O and Av = O (respectively). We need

to show that A(ru + sv) = O. Clearly,

A(ru + sv) = A(ru) + A(sv) = r(Au) + s(Av) = O + O = O.

Therefore, ru + sv is also a solution to the system.


Example 1.6.13

(T #12pg.89

)

Let u and v be two solutions of the non-homogenous system Ax = B. Show that u− v

is a solution to the homogenous system Ax = O.

Solution:

Since u and v are two solution of Ax = B, then Au = B and Av = B (respectively). We need

to show that A(u− v) = O. Clearly,

A(u− v) = Au− Av = B −B = O.

Therefore, u− v is a solution to the system Ax = O.


Exercise 1.6.1

1. Let A =

1

3

−1

3

. Find all constants c ∈ R such that (cA)T · (cA) = 5.

2. Find the reduced row echelon form (r.r.e.f.) of the following matrix:

2 4 6 0

1 2 4 0

1 3 3 1

3. Let A =

1 0 3 3

0 1 1 −1

1 −2 3 1

0 2 0 a2 + 1

and B =

1

0

0

a + 2

. Find all value(s) of a such that the

system Ax = B has at least one solution.

4. Solve the system

x + y + z = 0

x + 2y + 3z = 0

x + 3y + 4z = 0

x + 4y + 5z = 0.

5. Find all value(s) of a for which the system

x − y + (a + 3)z = a3 − a− 7

−x + ay − az = a

2(a− 1)y + (a2 + 2)z = 8a− 14.

has (a) no solution, (b) unique solution, and (c) infinite many solutions.

6. Show that if C1 and C2 are solutions of the system Ax = B, then 4C1 − 3C2 is also a

solution of this system.

1.7. The Inverse of a Matrix 31

1.7 The Inverse of a Matrix

Unless otherwise specified, all matrices in this section are considered to be square matrices.

Definition 1.7.1

A matrix A ∈ Mn×n is called non-singular or invertable if there exists a matrix B ∈ Mn×n

such that

A B = B A = In.

In particular, B is called the inverse of A and is denoted by A−1. If there is no such B, we

say that A is singular which means that A has no inverse.

Example 1.7.1

Here is an example of a 2× 2 matrix along with its inverse.

1 1

2 3

3 −1

−2 1

=

3 −1

−2 1

1 1

2 3

=

1 0

0 1

= I2.

Theorem 1.7.1

If a matrix has an inverse, then its inverse is unique.

Proof:

Assume that A ∈Mn×n is a matrix with two inverses B and C, then

B = B In = B (A C) = (B A) C = In C = C.

Therefore, A has a unique inverse.

Remark 1.7.1

⋆ How to find A−1 for a given A ∈Mn×n:

1. form the augmented matrix [ A | In ],


2. find the r.r.e.f. of [ A | In ], say [ C |B ]:

(a) if C = In, then A is non-singular and A−1 = B,(b) if C 6= In, then C has a row of zeros and A is singular matrix with no inverse.

Example 1.7.2

Find, if possible, A−1 for A =

1 1

2 3

Solution:

1 1 1 0

2 3 0 1

r2−2r1→r2−−−−−−→

1 1 1 0

0 1 −2 1

r1−r2→r1−−−−−−→

1 0 3 −1

0 1 −2 1

.

Therefore, A is non-singular and A−1 =

3 −1

−2 1

.

Example 1.7.3


1 0 1

1 1 2

2 0 1

Solution:

1 0 1 1 0 0

1 1 2 0 1 0

2 0 1 0 0 1

r2−r1→r2−−−−−−→r3−2r1→r3

1 0 1 1 0 0

0 1 1 −1 1 0

0 0 −1 −2 0 1

−r3−−→

1 0 1 1 0 0

0 1 1 −1 1 0

0 0 1 2 0 −1

r1−r3→r1−−−−−−→r2−r3→r2

1 0 0 −1 0 1

0 1 0 −3 1 1

0 0 1 2 0 −1

.

Therefore, A is non-singular and A−1 =

−1 0 1

−3 1 1

2 0 −1

.


Example 1.7.4


1 2 −3

1 −2 1

5 −2 −3

Solution:

1 2 −3 1 0 0

1 −2 1 0 1 0

5 −2 −3 0 0 1

r2−r1→r2−−−−−−→r3−5r1→r3

1 2 −3 1 0 0

0 −4 4 −1 1 0

0 −12 12 −5 0 1

r3−3r2→r3−−−−−−→

1 2 −3 1 0 0

0 −4 4 −1 1 0

0 0 0 −2 −3 1

.

At this point, we can conclude that this matrix is singular with no inverse because of the fact

that the third row in the first part is a zero row. In particular, A−1 does not exist.

Remark 1.7.2

Let A be an n× n matrix:

1. if A is row equivalent to In, then it is non-singular,

Example 1.7.5

A =

−3 0

0 5

is row equivalent to I2.

2. if A is row equivalent to a matrix with a row of zeros, then it is singular.

Example 1.7.6

A =

−1 1

0 0

is row equivalent to a matrix with a row of zeros.

TRUE or FALSE:


⋆ If A and B are singular matrices, then so is A + B. (FALSE).

reason: A =

1 0

0 0

and B =

0 0

0 1

are two singular matrices, while A + B =

1 0

0 1

is a non-

singular.

⋆ If A and B are non-singular matrices, then so is A + B. (FALSE).

reason: A =

1 0

0 1

and B =

−1 0

0 −1

are two non-singular matrices, while A + B =

0 0

0 0

is a

singular.

Theorem 1.7.2

If A is non-singular, then A−1 is non-singular and (A−1)−1 = A.

Proof:

If A is non-singular, then A A−1 = A−1 A = I, or A−1 A = A A−1 = I. Therefore, A−1 is

non-singular and (A−1)−1 = A.

Theorem 1.7.3

If A and B are non-singular matrices, then A B is non-singular and (A B)−1 = B−1 A−1.

Proof:

Since both A and B are non-singular, then both A−1 and B−1 exist. Thus

A B (B−1 A−1) = A (B B−1) A−1 = A I A−1 = A A−1 = I.

A B (B−1 A−1) = A (B B−1) A−1 = A I A−1 = A A−1 = I.

Thus, A B is non-singular and (A B)−1 = B−1 A−1.

Theorem 1.7.4

If A is non-singular, then AT is non-singular and (AT )−1 = (A−1)T .

Proof:

A is non-singular, and thus A A−1 = I = A−1 A. Taking the transpose for the whole thing, we

get


(A A−1)T = I = (A−1 A)T }

⇒(A−1)T AT = I = AT (A−1)T

AT is non-singular and (AT )−1 = (A−1)T .

⋆ Inverses and System of Linear Equations

Remark 1.7.3

Assume that A is a non-singular n× n matrix. Then the non-homogenous system Ax = B has

a unique solution since

A−1 A x = A−1 B

In x = A−1 B

x = A−1 B

Remark 1.7.4

Assume that A is a non-singular n × n matrix. Then the homogenous system Ax = O has a

trivial solution since

A−1 A x = A−1 O

x = O

Remark 1.7.5

Assume that A is a singular n×n matrix. Then the homogenous system Ax = O has a non-trivial

solution.

Example 1.7.7

Solve the following system ”A x = B”

1 0 1

1 1 2

2 0 1

x

y

z

=

0

1

−1

Solution:


We can find the inverse of A as in Example 1.7.3, to get

A−1 =

−1 0 1

−3 1 1

2 0 −1

Then, using Remark 1.7.3,

x =

x

y

z

= A−1 B =

−1 0 1

−3 1 1

2 0 −1

0

1

−1

=

−1

0

1

which is a unique solution.

Example 1.7.8

Solve the following system ”A x = O”

1 0 1

1 1 2

2 0 1

x

y

z

=

0

0

0

Solution:

We can find the inverse of A as in Example 1.7.3, to get

A−1 =

−1 0 1

−3 1 1

2 0 −1

Then, using Remark 1.7.4,

x =

x

y

z

= A−1 O =

−1 0 1

−3 1 1

2 0 −1

0

0

0

=

0

0

0

which is a trivial solution.


Example 1.7.9

Solve the following system ”A x = O”

1 2 −3

1 −2 1

5 −2 −3

x

y

z

=

0

0

0

Solution:

Since A is singular as we have seen in Example 1.7.4, we can not follow Remark 1.7.4. Therefore,

we use the usual Guass-Jordan method to solve this system, knowing that this system has

nontrivial solution as in Remark 1.7.5.

1 2 −3 0

1 −2 1 0

5 −2 −3 0

≈ · · · ≈

1 0 −1 0

0 1 −1 0

0 0 0 0

Thus, the reduced system is

x − z = 0

y − z = 0

. Let z = t ∈ R. Therefore, x =

t

t

t

Remark 1.7.6

For any A ∈Mn×n, we have

1. A is non-singular ⇐⇒ A ≈ In ⇐⇒ A−1 exits ⇐⇒ A x = B has a unique solution ⇐⇒A x = O has a trivial solution.

2. A is singular⇐⇒ A ≈ a matrix with a row of zeros⇐⇒ A−1 Does Not Exist⇐⇒ A x = O

has a non-trivial solution.


Exercise 1.7.1

1. Let AT =

1 0 1

1 1 0

2 0 1

. Find (2A)−1.

[

Hint: If c is a nonzero constant, then what is (cA)−1?]

2. Let A =

1 2

1 3

, and B =

1 −1 2

3 2 0

. Find C if (BT + C)A−1 = BT .

3. Let A−1 =

1 2 0

0 1 −1

2 0 1

and B =

1 1 0

1 0 0

0 1 1

. Find C if A C = BT .

[

Hint: Consider multiplying both sides from the left with A−1.]

4. Let A−1 =

1 2 3

0 1 2

0 0 1

. Find all x, y, z ∈ R such that

[

x y z

]

A =[

1 2 3

]

.

5. Let A =

0 1 −1

1 1 0

0 1 2

and B =

1 2 −1

1 3 0

1 1 −1

.

(a) find B−1.

(b) Find C if A = B C.

6. Let A, B, and C be n× n matrices such that D = AB + AC is non-singular.

(a) Find A−1 if possible.

(b) Find (B + C)−1 if possible.

7. Let A be an n× n matrix such that A3 = O. Prove or disprove that

(In − A)−1 = A2 + A + In.


CHAPTER 1 REVISION

Revision of Chapter 1 ... Have Fun :-)

1. Let A =

5 −10

a −4

. Find all values of a such that A2 = A.

2. Find the matrix A satisfying

2AT − 3

1 2

−1 1

T

=

2 3

−1 2

.

3. Let A =

1 −3

−2 4

and X =

5

3

. Compute, if possible, XT AT , AT XT , XT X, and XXT .

4. Let A =

1 1

0 2

2 0

and B =

4 1 −1

2 5 1

.

(a) Find, if possible, AB and AT B.

(b) Express the second column of AB as a linear combination of the columns of A.

(c) Express the second row of AB as a linear combination of the rows of B.

5. ⋆ Let A be a square matrix. Show that if A = 2AT , then A = O.

6. Let A =

1 −1 2

2 1 1

1 1 0

.

(a) Find all solution, if any, to the system A

x

y

z

=

1

8

7

.

(b) Find all solution to the homogenous system A

x

y

z

=

0

0

0

.


7. Find all values of a such that the following system has at least one solution.

x + y + z = 1

x + 3y + 2z = 2

x + y + (a2 − 3)z = a− 1

.

8. Let A =

1 2 −1

1 3 0

1 1 −1

, B =

1 3

2 0

1 −2

, and C =

2 −1

1 1

0 3

.

(a) Find A−1.

(b) Find a matrix X such that AX + B = C.

(c) Is it possible to find a matrix Y such that Y A + B = C? Explain.

9. Let A and B be skew symmetric matrices. Show that AB is symmetric iff AB = BA.

10. Let A and B be two square matrices such that AB = O. If B is non-singular, find A.

11. Find a 2×2 matrix C such that AT CB = I2, where A−1 =

−1 1

1 0

and B−1 =

2 −1

0 1

.

12. Let P =

1 0 2

1 1 0

1 1 1

and Q =

1 2 −2

−1 −1 2

0 −1 1

.

(a) Compute PQ.

(b) Is P singular or non-singular matrix? Explain.

13. ⋆ Consider the system

3x +2y −z −w = 0

−x −y +3z −5w = 0

5y +2z −4w = 0

x −y −z +w = 0

,

(a) Determine whether (1, 0, 2, 1) is a solution to the above system.

(b) Does the above system has infinitely many solutions? Explain.

(c) Let A be the coefficient matrix of the above system. Is A singular? Explain.


14. Let A−1 =

1 2 3

0 1 2

0 0 1

. Find all x, y, z ∈ R such that

[

x y z

]

A =[

1 2 3

]

.

15. Let A, B, and C be n× n matrices such that D = AB + AC is non-singular.

(a) Find A−1 if possible.

(b) Find (B + C)−1 if possible.

16. Let A be an n× n matrix such that A3 = O. Prove or disprove that

(In − A)−1 = A2 + A + In.

17. ⋆ Let A be an n× n matrix and let U and V be two n× 1 column matrices. Show that if

AU = AV and U 6= V , then A is singular.

18. Solve the following linear system

x +3y −z +w = 1

2x −y −2z +2w = 2

3x +y −z +w = 1

.

19. Let A =

1 2 4

3 −2 0

4 3 9

and C =

2

−5

1

.

(a) If possible, compute (AC + I3)T .

(b) Write AC as a linear combination of the columns of A.

(c) Find a 3× 3 matrix B, if any, such that A−1B + I3 = A.

20. Let A =

1 2

1 3

and B =

1 −1 2

3 2 0

.

(a) If possible, find A2 − (3A)T .

(b) Find C, if any, such that (BT + C)A−1 = BT .

2

Chapter

Determinants

2.1 Determinants

Definition 2.1.1

Let S = {1, 2, · · · , n} be a set of integers from 1 to n. A rearrangement of j1, j2, · · · , jn of

elements of S is called permutation of S.

Example 2.1.1

For S = {1, 2}, we have 2 permutations:

1 2, and 2 1.

While, for S = {1, 2, 3}, we have the following 6 permutations:

1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, and 3 2 1.

In general, for S = {1, 2, · · · , n}, we have n! = n(n− 1)(n− 2) · · · 3 · 2 · 1 permutations.

Definition 2.1.2

A permutation j1j2 · · · jn of the set S = {1, 2, · · · , n} is said to have an inversion if a larger

integer jr preceedes a smaller one js for r, s ∈ S. A permutation is called even with a positive

”+” sign or odd with a negative ”−” sign according to whether the total number of inversions

in it is even or odd, respectively.

43

44 Chapter 2. Determinants

Example 2.1.2

If S = {1, 2, 3}, then

permutation #inversions even-odd sign inversions

1 2 3 0 even + 1 2 1 3 2 3

1 3 2 1 odd − 1 3 1 2 3 2

2 1 3 1 odd − 2 1 2 3 1 3

2 3 1 2 even + 2 3 2 1 3 1

3 1 2 2 even + 3 1 3 2 1 2

3 2 1 3 odd − 3 2 3 1 2 1

TRUE or FALSE:

⋆ The permutation 5 2 1 3 4 has a positive sign. (FALSE).

reason: The number of inversions is 5 which are 5 2, 5 1, 5 3, 5 4, and 2 1. So, this permutation has an

odd number of inversions and a negative sign.

Definition 2.1.3

Let A = (aij) ∈Mn×n. Then, the determinant of A, denoted by det(A) or∣∣∣ A

∣∣∣, is

det(A) =∑

(±)a1j1a2j2a3j3 · · · anjn

where the summation ranges over all permutations j1j2 · · · jn of the set S = {1, 2, · · · , n}. The

sign is taken as + or − according to the sign of the permutation.

Example 2.1.3

Compute the determinant of A =

a11 a12

a21 a22

.

Solution:

Using the definition, we have det(A) =∑

(±)a1j1a2j2 , where j1j2 is a permutation of S = {1, 2}.Thus, j1j2 ∈ {1 2, 2 1} and

det(A) = + a11a22 − a12a21.

2.1. Determinants 45

Example 2.1.4

Compute the determinant of A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

.

Solution:

Using the definition, we have det(A) =∑

(±)a1j1a2j2a3j3 , where j1j2j3 is a permutation of S =

{1, 2, 3}. Thus, j1j2j3 ∈ {1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1} and

det(A) = + a11a22a33 − a11a23a32 − a12a21a33 + a12a23a31 + a13a21a32 − a13a22a31.

Moreover, this formula can be found by taking the sum of the positive product of the diagonal

entries and the negative product of the anti-diagonal entries in the following matrix:

+ + +− − −a11 a12 a13 a11 a12

a21 a22 a23 a21 a22

a31 a32 a33 a31 a32

Example 2.1.5

Find det(A) for

A =

2 3

1 5

.

Solution:

∣∣∣ A

∣∣∣ = (2)(5)− (1)(3) = 10− 3 = 7.


Example 2.1.6

Find det(A) for

A =

2 0 1

−2 3 4

5 1 2

.

Solution:

We calculate the det(A) using the following diagram:

+ + +− − −2 0 1 2 0

−2 3 4 −2 3

5 1 2 5 1

In particular,

∣∣∣ A

∣∣∣ = (2)(3)(2) + (0)(4)(5) + (1)(−2)(1)

−(1)(3)(2)− (2)(4)(1)− (0)(−2)(2) = 12 + 0 + (−2)− 15− 8− 0 = −13.

⋆ Properties of Determinants:

1.∣∣∣ A

∣∣∣ =

∣∣∣ AT

∣∣∣.

2. If B is obtained from A by interchanging two rows (or two columns), then∣∣∣ B

∣∣∣ = −

∣∣∣ A

∣∣∣.

3. If A has two equal rows (or columns), then∣∣∣ A

∣∣∣ = 0.

4. If A has a row (or a column) of zeros, then∣∣∣ A

∣∣∣ = 0.

5. If B is obtained from A by adding a multiple of a row (or a column) to another row (or column,

respectively), then∣∣∣ A

∣∣∣ =

∣∣∣ B

∣∣∣.

6. If B is obtained from A by multiplying a row (or a column) by a scalar c, then∣∣∣ B

∣∣∣ = c

∣∣∣ A

∣∣∣.


Remark 2.1.1

In general, for A ∈Mn×n, we have∣∣∣ c A

∣∣∣ = cn

∣∣∣ A

∣∣∣.

7. The determinant of lower triangular, upper triangular, diagonal, and scalar matrices can be com-

puted as follows.

Theorem 2.1.1

The determinant of lower triangular, upper triangular, diagonal, and scalar matrices equal

the product of elements of the diagonal.

8. For any A, B ∈Mn×n, we have∣∣∣ A B

∣∣∣ =

∣∣∣ A

∣∣∣

∣∣∣ B

∣∣∣.

9.

Theorem 2.1.2

If A is an n × n non-singular matrix, then∣∣∣ A

∣∣∣ 6= 0 and

∣∣∣ A−1

∣∣∣ = 1

| A |. This statement

suggests that if∣∣∣ A

∣∣∣ = 0, then A is singular matrix.

Proof:

Since A is non-singular, A−1 exists, then

AA−1 = In (take the determinant for both sides)∣∣∣A A−1

∣∣∣ =

∣∣∣ In

∣∣∣

∣∣∣ A

∣∣∣

∣∣∣ A−1

∣∣∣ = 1.

Thus,∣∣∣ A

∣∣∣ 6= 0 and

∣∣∣ A−1

∣∣∣ = 1

| A |.

Example 2.1.7

Let A, B ∈Mn×n. Show that, if A B = In, then B A = In.

Solution:


If A B = In, then∣∣∣ A

∣∣∣ 6= 0 and

∣∣∣ B

∣∣∣ 6= 0.

A−1 A B = A−1 In

In B = A−1

B A = A−1 A = In

Thus, A B = In = B A.

Example 2.1.8

(T #16pg.195

)

Show that if A ∈Mn×n is skew-symmetric matrix and n is odd, then∣∣∣ A

∣∣∣ = 0.

Solution:

Since A is skew-symmetric, then AT = −A and taking the determinant for both sides

∣∣∣ AT

∣∣∣ =

∣∣∣ − A

∣∣∣

∣∣∣ AT

∣∣∣ = (−1)n

∣∣∣ A

∣∣∣, where n is odd and (−1)n = −1.

∣∣∣ A

∣∣∣ =

∣∣∣ AT

∣∣∣ = −

∣∣∣ A

∣∣∣.

Therefore,∣∣∣ A

∣∣∣ = −

∣∣∣ A

∣∣∣ which means that

∣∣∣ A

∣∣∣ = 0.

TRUE or FALSE:

⋆ If A, B ∈Mn×n with∣∣∣ A

∣∣∣ =

∣∣∣ B

∣∣∣, then A = B. (FALSE).

reason: I2 6= −I2 while∣∣∣ I2

∣∣∣ = 1 and

∣∣∣ − I2

∣∣∣ = (−1)2

∣∣∣ I2

∣∣∣ = 1.

Example 2.1.9

(T #5

pg.194

)

If |A B| = 0, then either∣∣∣ A

∣∣∣ = 0 or

∣∣∣ B

∣∣∣ = 0.

Solution:

Clearly, |A B| =∣∣∣ A

∣∣∣

∣∣∣ B

∣∣∣ = 0. Then,

∣∣∣ A

∣∣∣ = 0 or

∣∣∣ B

∣∣∣ = 0.


Example 2.1.10

(T #8

pg.194

)

If A B = In, then∣∣∣ A

∣∣∣ 6= 0 and

∣∣∣ B

∣∣∣ 6= 0.

Solution:

|A B| = |In| =⇒ |A B| = 1 which implies that∣∣∣ A

∣∣∣

∣∣∣ B

∣∣∣ = 1 and both

∣∣∣ A

∣∣∣ and

∣∣∣ B

∣∣∣ can not be

zero.

Example 2.1.11

(T #9(b)pg.194

)

Show that if A−1 = AT , then∣∣∣ A

∣∣∣ = 1 or

∣∣∣ A

∣∣∣ = −1.

Solution:

Since A−1 = AT , we have

|A−1| = |AT | ⇐⇒ 1

|A| =∣∣∣ A

∣∣∣⇐⇒

∣∣∣ A

∣∣∣

2= 1⇐⇒

∣∣∣ A

∣∣∣ = ±1.

Example 2.1.12

(T #10pg.194

)

Show that if A is non-singular and A2 = A, then∣∣∣ A

∣∣∣ = 1.

Solution:

Since A is non-singular, A−1 exists and

A2 = A⇐⇒ A = In ⇐⇒∣∣∣ A

∣∣∣ = |In| ⇐⇒

∣∣∣ A

∣∣∣ = 1.

Example 2.1.13

(T #10pg.210

)

Show that for any A, B, C ∈Mn×n, if A B = A C and∣∣∣ A

∣∣∣ 6= 0, then B = C.

Solution:

Since∣∣∣ A

∣∣∣ 6= 0, then A−1 exists and thus

AB = AC ⇐⇒ A−1AB = A−1AC ⇐⇒ B = C.


Example 2.1.14

Let A, B ∈M3×3 with∣∣∣ A

∣∣∣ = 2 and

∣∣∣ B

∣∣∣ = −2. Find

∣∣∣ 2 A−1 (B2)T

∣∣∣.

Solution:

∣∣∣ 2 A−1 (B2)T

∣∣∣ = 23

∣∣∣ A−1

∣∣∣

∣∣∣ B2

∣∣∣ = 8

1

|A|∣∣∣ B

∣∣∣

∣∣∣ B

∣∣∣

= 8 (1

2) (−2)(−2) = 16.

Example 2.1.15

Evaluate the determinant of the matrix A via reduction to triangular form , where

A =

2 −2 5

0 3 1

1 4 2

.

Solution:

The idea here is to compute the determinant of A by applying the method of Theorem 2.1.1.

∣∣∣ A

∣∣∣ =

∣∣∣∣∣∣∣∣∣

2 −2 5

0 3 1

1 4 2

∣∣∣∣∣∣∣∣∣

r1↔r3−−−→ (−1)

∣∣∣∣∣∣∣∣∣

1 4 2

0 3 1

2 −2 5

∣∣∣∣∣∣∣∣∣

r3−2r1→r3−−−−−−→ (−1)

∣∣∣∣∣∣∣∣∣

1 4 2

0 3 1

0 −10 1

∣∣∣∣∣∣∣∣∣

c2↔c3−−−→

(−1)(−1)︸︷︷︸

=+1

∣∣∣∣∣∣∣∣∣

1 2 4

0 1 3

0 1 −10

∣∣∣∣∣∣∣∣∣

r3−r2→r3−−−−−−→

∣∣∣∣∣∣∣∣∣

1 2 4

0 1 3

0 0 -13

∣∣∣∣∣∣∣∣∣

= (1)(1)(−13) = −13.

Example 2.1.16

Let

∣∣∣∣∣∣∣∣∣

a1 b1 c1

a2 b2 c2

a3 b3 c3

∣∣∣∣∣∣∣∣∣

= 3. Find

∣∣∣∣∣∣∣∣∣

a1 a2 a3

−2b1 −2b2 −2b3

3a1 − c1 3a2 − c2 3a3 − c3

∣∣∣∣∣∣∣∣∣

.

Solution:

We solve this question in the following two ways (either one works):


⋆ solution 1:∣∣∣∣∣∣∣∣∣

a1 b1 c1

a2 b2 c2

a3 b3 c3

∣∣∣∣∣∣∣∣∣

= 3transpose−−−−−→

∣∣∣∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣∣∣

= 3(−2)r2→r2−−−−−−→(−1)r3→r3

∣∣∣∣∣∣∣∣∣

a1 a2 a3

−2b1 −2b2 −2b3

−c1 −c2 −c3

∣∣∣∣∣∣∣∣∣

=

(−1)(−2)3r3+3r1→r3−−−−−−→

∣∣∣∣∣∣∣∣∣

a1 a2 a3

−2b1 −2b2 −2b3

3a1 − c1 3a2 − c2 3a3 − c3

∣∣∣∣∣∣∣∣∣

= 6.

⋆ solution 2:∣∣∣∣∣∣∣∣∣

a1 a2 a3

−2b1 −2b2 −2b3

3a1 − c1 3a2 − c2 3a3 − c3

∣∣∣∣∣∣∣∣∣

= (−2)(−1)

∣∣∣∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 − 3a1 c2 − 3a2 c3 − 3a3

∣∣∣∣∣∣∣∣∣

r3+3r1→r3−−−−−−→

2

∣∣∣∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣∣∣

transpose−−−−−→ 2

∣∣∣∣∣∣∣∣∣

a1 b1 c1

a2 b2 c2

a3 b3 c3

∣∣∣∣∣∣∣∣∣

= 2(3) = 6.


Exercise 2.1.1

1. Let D =

0 4 4 0

1 1 2 0

1 3 5 3

0 1 2 6

. Evaluate∣∣∣D

∣∣∣.

[

Hint: |D| = −12.]

2. Let A =

a + e b + f

c d

, B =

a b

c d

, and C =

e f

c d

. Show that∣∣∣ A

∣∣∣ =

∣∣∣ B

∣∣∣ +

∣∣∣ C

∣∣∣.

3. Given that

∣∣∣∣∣∣∣∣∣

a1 b1 c1

a2 b2 c2

a3 b3 c3

∣∣∣∣∣∣∣∣∣

= 7, evaluate

∣∣∣∣∣∣∣∣∣

b1 b2 b1 − 3b3

a1 a2 a1 − 3a3

c1 c2 c1 − 3c3

∣∣∣∣∣∣∣∣∣

.

[

Hint: The result is 21.]

4. Let A =

a1 a2 a3

b1 b2 b3

c1 c2 c3

, and B =

2a3 2a2 2a1

b3 − a3 b2 − a2 b1 − a1

c3 + 3b3 c2 + 3b2 c1 + 3b1

. If

∣∣∣A

∣∣∣ = −4, find

∣∣∣B

∣∣∣.

[

Hint: The result is 8.]

5. Let A be a matrix with A−1 =

7 1 0 3

2 0 0 0

1 3 5 4

6 2 0 5

. Find det (A).

6. Find all values of α for which the matrix

1 2 4

1 3 9

1 α α2

is singular.

7. Let A and B be two n×n matrices such that A is invertible and B is singular. Prove that

A−1B is singular.

8. If A and B are 2× 2 matrices with det (A) = 2 and det (B) = 5, compute∣∣∣3 A2(AB−1)T

∣∣∣.

2.2. Cofactor Expansion and Applications 53

2.2 Cofactor Expansion and Applications

Definition 2.2.1

Let A = (aij) be an n× n matrix. LetMij be the (n− 1)× (n− 1) submatrix of A obtained by

deleting the ith row and the jth column of A. The determinant, det(Mij), is called the minor

of aij. The cofactor denoted by Aij of aij is defined by

Aij = (−1)i+j det(Mij).

Moreover, the cofactor matrix denoted by coef(A) = (Aij) where 1 ≤ i, j ≤ n.

Example 2.2.1

Compute the minors, cofactors, and the coef(A) for A =

1 2 1

0 1 −1

3 1 −2

.

Solution:

minors: cofactors:

det(M11) =

∣∣∣∣∣∣

1 −1

1 −2

∣∣∣∣∣∣

= -1 =⇒ A11 = (−1)2(−1) = -1

det(M12) =

∣∣∣∣∣∣

0 −1

3 −2

∣∣∣∣∣∣

= 3 =⇒ A12 = (−1)3(3) = -3

det(M13) =

∣∣∣∣∣∣

0 1

3 1

∣∣∣∣∣∣

= -3 =⇒ A13 = (−1)4(−3) = -3

det(M21) =

∣∣∣∣∣∣

2 1

1 −2

∣∣∣∣∣∣

= -5 =⇒ A21 = (−1)3(−5) = 5

det(M22) =

∣∣∣∣∣∣

1 1

3 −2

∣∣∣∣∣∣

= -5 =⇒ A22 = (−1)4(−5) = -5

det(M23) =

∣∣∣∣∣∣

1 2

3 1

∣∣∣∣∣∣

= -5 =⇒ A23 = (−1)5(−5) = 5


det(M31) =

∣∣∣∣∣∣

2 1

1 −1

∣∣∣∣∣∣

= -3 =⇒ A31 = (−1)4(−3) = -3

det(M32) =

∣∣∣∣∣∣

1 1

0 −1

∣∣∣∣∣∣

= -1 =⇒ A32 = (−1)5(−1) = 1

det(M33) =

∣∣∣∣∣∣

1 2

0 1

∣∣∣∣∣∣

= 1 =⇒ A33 = (−1)6(1) = 1

Therefore, coef(A) =

−1 −3 −3

5 −5 5

−3 1 1

.

Theorem 2.2.1

Let A = (aij) be an n× n matrix with cofactors Aij for all 1 ≤ i, j ≤ n, then for each 1 ≤ i ≤ n

we have

ai1Ak1 + ai2Ak2 + · · ·+ ainAkn =

∣∣∣ A

∣∣∣ if i = k,

0 if i 6= k.

and for each 1 ≤ j ≤ n

a1jA1k + a2jA2k + · · ·+ anjAnk =

∣∣∣ A

∣∣∣ if j = k,

0 if j 6= k.

Example 2.2.2

Compute the determinant of A by using the cofactor expansion method, where A is given in

Example 2.2.1.

Solution:

Computing the determinant of A using the first row of A, we have

∣∣∣ A

∣∣∣ = a11A11 + a12A12 + a13A13 = (1)(−1) + (2)(−3) + (1)(−3) = −10.

Moreover, we can get the same result using (for instance) the second column of A,

∣∣∣ A

∣∣∣ = a12A12 + a22A22 + a32A32 = (2)(−3) + (1)(−5) + (1)(1) = −10.


Note that, if we choose the first row of A and the cofactors of (for instance) the second row of

A, we get

a11A21 + a12A22 + a13A23 = (1)(5) + (2)(−5) + (1)(5) = 0.

Example 2.2.3

Compute

∣∣∣∣∣∣∣∣∣∣∣∣

0 3 0 1

2 1 1 2

0 0 1 2

0 1 0 1

∣∣∣∣∣∣∣∣∣∣∣∣

using the cofactor expansion method.

Solution:

We solve this problem by choosing the row or the column with the maximum number of zeros

as this will ease the computations: Choosing the first column, we have

∣∣∣ A

∣∣∣ = a11A11 + a21A21 + a31A31 + a41A41

= ✘✘✘✘(0)A11 +✘✘✘✘(0)A21 +

∣∣∣∣∣∣∣∣∣∣∣∣

0 3 0 1−

2 1 1 2

0 0 1 2

0 1 0 1

∣∣∣∣∣∣∣∣∣∣∣∣

+✘✘✘✘(0)A41 = (−2)

choose 2nd column︷︸︸︷∣∣∣∣∣∣∣∣∣

3 0 1

0 1 2

1 0 1

∣∣∣∣∣∣∣∣∣

= (−2)

=O︷︸︸︷∣∣∣∣∣∣∣∣∣∣

3−

0 1

0 1 2

1 0 1

∣∣∣∣∣∣∣∣∣∣

+

∣∣∣∣∣∣∣∣∣∣

3 0 1

0+

1 2

1 0 1

∣∣∣∣∣∣∣∣∣∣

+

=O︷︸︸︷∣∣∣∣∣∣∣∣∣∣

3 0 1

0 1 2

1−

0 1

∣∣∣∣∣∣∣∣∣∣

= (−2)

(+1)

∣∣∣∣∣∣∣

3 1

1 1

∣∣∣∣∣∣∣

= (−2) [3− 1] = (−2)(2) = −4.


Definition 2.2.2

Let A = (aij) ∈Mn×n. The adjoint matrix of A, denoted by adj(A), is given by

adj(A) = (coef(A))T .

Example 2.2.4

Compute the adjoint of A, where A as given in Example 2.2.1:

A =

1 2 1

0 1 −1

3 1 −2

.

Solution:

Here, we use the result of Example 2.2.1, where

coef(A) =

−1 −3 −3

5 −5 5

−3 1 1

.

Therefore,

adj(A) = (coef(A))T =

−1 5 −3

−3 −5 1

−3 5 1

.

Theorem 2.2.2

Let A = (aij) ∈Mn×n. Then,

A · adj(A) = adj(A) · A =∣∣∣ A

∣∣∣ In.

Proof:


We first prove that A · adj(A) =∣∣∣ A

∣∣∣ · In,

A · adj(A) =

a11 a12 · · · a1n

a21 a22 · · · a2n

......

. . ....

an1 an2 · · · ann

A11 A21 · · · An1

A12 A22 · · · An2

......

. . ....

A1n A2n · · · Ann

=

∣∣∣ A

∣∣∣ 0 0 · · · 0

0∣∣∣ A

∣∣∣ 0 · · · 0

0 0. . . . . .

......

.... . . . . . 0

0 0 · · · 0∣∣∣ A

∣∣∣

=∣∣∣ A

∣∣∣

1

. . . 01

0 . . .

1

=∣∣∣ A

∣∣∣ · In.

The proof of adj(A) · A =∣∣∣ A

∣∣∣ · In is similar.

Example 2.2.5

Let adj(A) =

1 2

4 6

. Find A.

Solution:

Assume that A =

x y

z w

. Then coef(A) =

w −z

−y x

. Therefore,

adj(A) = (coef(A))T =

w −y

−z x

=

1 2

4 6

.

Thus, x = 6, y = −2, z = −4, and w = 1, and A =

6 −2

−4 1


Theorem 2.2.3

If A is non-singular, then A−1 = 1|A|

adj(A).

Proof:

A is non-singular so that∣∣∣ A

∣∣∣ 6= 0 and A−1 exists. Then, by Theorem 2.2.2,

A adj(A) =∣∣∣ A

∣∣∣ In multiply from left by A−1

A−1 · A adj(A) = A−1 ·∣∣∣ A

∣∣∣ In

adj(A) =∣∣∣ A

∣∣∣A−1

A−1 =1

|A| adj(A).

Remark 2.2.1

Let A =

a b

c d

be any 2× 2 non-singular matrix. Thus, ad− bc 6= 0, and

A−1 =1

|A|adj(A) =1

ad− bc

d −b

−c a

.

Confirmation:

A · A−1 =1

ad− bc

ad− bc −ab + ab

cd− cd −cb + ad

=

1 0

0 1

= I2.

Example 2.2.6

Find the inverse of A =

1 2

4 6

.

Solution:

By Remark 2.2.1, we have A−1 = −12

6 −2

−4 1

. The reader can check the solution by confirm-

ing that either (which leads to both) A A−1 = I2 or A−1 A = I2.


Example 2.2.7

(T #8

pg.210

)

Show that if A is a non-singular n× n matrix, then

∣∣∣ adj(A)

∣∣∣ =

∣∣∣ A

∣∣∣

n−1.

Solution:

Given that∣∣∣ A

∣∣∣ 6= 0, we use Theorem 2.2.2 to get

A adj(A) = |A| In∣∣∣A adj(A)

∣∣∣ =

∣∣∣|A| In

∣∣∣

∣∣∣ A

∣∣∣

∣∣∣ adj(A)

∣∣∣ =

∣∣∣ A

∣∣∣

n |In|∣∣∣ adj(A)

∣∣∣ =

∣∣∣ A

∣∣∣

n−1

Example 2.2.8

(T #12pg.210

)

Show that if A is non-singular, then adj(A) is non-singular and

(

adj(A))−1

=1

|A| A = adj(A−1).

Solution:

Since A is non-singular,∣∣∣ A

∣∣∣ 6= 0. We use Theorem 2.2.2 to get

adj(A) A = A adj(A) =∣∣∣ A

∣∣∣ In

(

adj(A)) 1

|A| A =1

|A| A(

adj(A))

= In

which shows that adj(A) is non-singular and

(

adj(A))−1

=1

|A| A. (2.2.1)

Now, if we replace A by A−1 in Theorem 2.2.2, we get

adj(A−1) A−1 = A−1 adj(A−1) =∣∣∣ A−1

∣∣∣ In.


Thus,

adj(A−1) · A−1 =1

|A| In

adj(A−1) · A−1 · A =1

|A| In · A

adj(A−1) =1

|A| A (2.2.2)

Adding Equation 2.2.1 and Equation 2.2.2, we get

(

adj(A))−1

=1

|A| A = adj(A−1).

Example 2.2.9

Let A be a non-singular 4× 4 matrix with∣∣∣ A−1

∣∣∣ = −2. Find

(a)∣∣∣ adj(A)

∣∣∣, and

(b)∣∣∣

12

AT adj(A−1)∣∣∣.

Solution:

We have∣∣∣ A

∣∣∣ = −1

2since

∣∣∣ A−1

∣∣∣ = −2. Then

(a):∣∣∣ adj(A)

∣∣∣ =

∣∣∣ A

∣∣∣

4−1=

∣∣∣ A

∣∣∣

3=

(−12

)3= −1

8.

(b):

∣∣∣∣

1

2AT adj(A−1)

∣∣∣∣ =

(1

2

)4 ∣∣∣AT

∣∣∣

∣∣∣adj(A−1)

∣∣∣

=1

24

∣∣∣ A

∣∣∣

∣∣∣ A−1

∣∣∣

3

=1

24· −1

2· (−2)3 =

1

4.


Example 2.2.10

Let A be a 4× 4 matrix with∣∣∣ adj(A)

∣∣∣ = −8. Find 2A adj(2A).

Solution:

By Theorem 2.2.2, we have

(2A) · adj(2A) = adj(2A) · (2A) = |2A| I4.

Since∣∣∣ adj(A)

∣∣∣ = −8, we have

∣∣∣ A

∣∣∣ = −2. Thus,

2A adj(2A) = |2A| I4

= (2)4∣∣∣ A

∣∣∣ I4

= (16) (−2) I4 = −32 I4.


Exercise 2.2.1

1. Let adj(A) =

2 8

1 4

. Find A.

2. Let A be a nonsingular n× n matrix.

(a) Show that adj(A) is nonsingular.

(b) Prove that adj(A−1) = 1det (A)

A.

3. Let A =

1 4 2

5 −3 6

2 3 2

. Compute the cofactors A11, A12, and A13, and show that 5A11 −

3A12 + 6A13 = 0.

4. Let A =

1 −1 2

0 3 1

−1 3 1

.

(a) Compute det (A) using a cofactor expansion along row 2.

(b) Find A adj(A).

(c) Find det (2A−1adj(A)).

5. Let A =

1 −1 2

0 3 1

−1 3 1

.

(a) Find |A| using a cofactor expansion along row 2.

(b) Find A adj(A).

(c) Find |2A−1adj(A)|.


CHAPTER 2 REVISION


1. Find all 3× 3 matrices A, if any, such that∣∣∣adj(A)

∣∣∣ = −3.

2. Let A and B be two 4 × 4 nonsingular matrices such that det (2A−1B2) = 32 and

det (B−1adj(A)) = 4. Find∣∣∣ A

∣∣∣ and

∣∣∣ B

∣∣∣.

3. Let A be a nonsingular n× n matrix. Prove that det (adj(A)) = [det (A)]n−1.

4. Show that if A is a nonsingular matrix, then adj(A) is nonsingular and

(adj(A))−1 =1

det (A)A = adj(A−1).

5. Let A =

1 2 −2 3

0 −1 3 1

0 2 −6 1

2 4 −3 0

. Find det(A) by:

(a) reducing A to triangular form.

(b) using a cofactor expansion of det(A).

(c) using the cofactor expansion on the fourth row of A.

6. Let A be a matrix with adj(A) = 12

0 0 6 0

3 0 7 0

1 4 8 0

7 1 9 6

. Find A−1.

7. Find adj(A), given that A−1 =

1 1 1 1

−1 7 6 4

4 4 4 0

0 1 1 1

.

8. Let A =

1 4 2

5 −3 6

2 3 2

. Compute the cofactors A11, A12, and A13, and show that 5A11 −

3A12 + 6A13 = 0.


9. Let A =

1 −1 2

0 3 1

−1 3 1

.

(a) Compute det (A) using a cofactor expansion along row 2.

(b) Find A adj(A).

(c) Find det (2A−1adj(A)).

10. Let A be an n × n non-singular matrix. Show that if A is symmetric, then adj(A) is

symmetric.

11. Let A =

1 −1 2

0 3 1

−1 3 1

.

(a) Find |A| using a cofactor expansion along row 2.

(b) Find A adj(A).

(c) Find |2A−1adj(A)|.

12. Answer each of the following as True or False (Justify your answer):

(a) If A and B are 2 × 2 matrices such that det (A) = −2 and det (B) = 8, then

det(

2AT adj(B−1))

= 1.

(b) if A and B are two 2×2 matrices such that det (adj(A)) = det (adj(B)), then A = B.

13. Let A =

1 0 2

2 1 1

0 1 1

.

(a) Find det(adj(A)).

(b) Find A2 adj(A).

14. Let A−1 =

1 2 3 1

1 4 6 0

0 2 0 0

2 3 2 3

.

(a) Find det(A) and A adj(A).

(b) Find adj(A).

3

Chapter

Vectors in Rn

3.2 Vectors in Rn

In this chapter, we deal with vectors in Rn, or sometimes we simply call them n-vectors. For instance,

x =[

x1 x2 · · · xn

]

, y =

y1

y2...

yn

, and z = 〈z1, z2, · · · , zn〉 are vectors in Rn.

Vecors in R3 can be manipulated by arrows starting at tail and pointing at head. For instance, vector

u = 〈3,−2, 4〉 and −u (which is of the same size in the opposite direction) both have their tail at the

origin O(0, 0, 0) and can be drawn as in the following figure (left). While vector v = 〈−1,−1, 1〉 has its

tail P (1, 2, 0) and its head Q(0, 1, 1), see the following figure (right).

x

y

z

u = 〈3, −2, 4〉

−u = 〈−3, 2, −4〉

3

−2

4O(0, 0, 0)

x

y

z

O(0, 0, 0)

P (1, 2, 0)

Q(0, 1, 1)

v = 〈−1, −1, 1〉

3.2.1 Vectors Operations:

Let x, y, z ∈ Rn and c, d ∈ R, then

65

66 Chapter 3. Vectors in Rn

1. Adding or substracting two vectors, they must have the same number of components:

x± y =

x1 ± y1

x2 ± y2...

xn ± yn

∈ Rn.

2. Multiplying a vector by a scalar:

c x =

c x1

c x2...

c xn

∈ Rn.

3. Rn is closed under vector addition (i.e. x + y ∈ R

n,∀x, y ∈ Rn):

(a) x + y = y + x,

(b) x + (y + z) = (x + y) + z,

(c) ∃ a vector O ∈ Rn such that x + O = O + x = x. ”additive identity”

(d) for any x ∈ Rn, ∃(−x) such that x + (−x) = (−x) + x = O. ”additive inverse”

4. Rn is closed under scalar multiplication (i.e. c x ∈ R

n,∀x ∈ Rn and c ∈ R):

(a) c (x + y) = c x + c y,

(b) (c + d)x = c x + d x,

(c) c (d x) = (c d) x,

(d) 1 x = x where 1 ∈ R.

3.2.2 The Dot Product of Vectors:

Definition 3.2.1

The dot (or inner or scalar) product of two vectors x = (x1, x2, · · · , xn) and y = (y1, y2, · · · , yn)

in Rn is

x · y =n∑

i=1

xi yi = x1y1 + x2y2 + · · ·+ xnyn.

3.2. Vectors in Rn 67

Definition 3.2.2

The length (or norm) of the vector x = (x1, x2, · · · , xn) ∈ Rn is

‖x‖ =√

x21 + x2

2 + · · ·+ x2n.

While, the distance between the points P (x1, x2, · · · , xn) and Q(y1, y2, · · · , yn) is the length of

the vector z =−→PQ = Q− P . That is,

‖z‖ =√

(x1 − y1)2 + (x2 − y2)2 + · · ·+ (xn − yn)2.

Remark 3.2.1

We note that, ‖x‖2 = x21 + x2

2 + · · ·+ x2n = x · x =⇒ ‖x‖2 = x · x.

Example 3.2.1

Let x =

2

4

3

1

and y =

5

3

2

1

be two vectors in R4. Find ‖y− x‖.

Solution:

Clearly, y− x =

3

−1

−1

0

=⇒ ‖y− x‖ =√

9 + 1 + 1 + 0 =√

11.


3.2.3 Properties of Dot Product:

Theorem 3.2.1

If x, y, z ∈ Rn and c ∈ R. Then,

1. x · x > 0 if x 6= 0 and x · x = 0 if x = 0,

2. x · y = y · x,

3. (x + y) · z = x · z + y · z,

4. (c x) · y = x · (c y) = c (x · y).

Example 3.2.2

(T #9

pg.246

)

If x ∈ Rn and c ∈ R, then ‖c x‖ = | c | ‖x‖.

Solution:

Let x = (x1, x2, · · · , xn) =⇒ c x = (c x1, c x2, · · · , c xn). Thus,

‖c x‖ =√

(c x1)2 + (c x2)2 + · · ·+ (cxn)2

=√

c2 (x21 + x2

2 + · · ·+ x2n)

= | c |√

x21 + x2

2 + · · ·+ x2n

= | c | ‖x‖.

3.2.4 The Angle Between Two Vectors

x

y

θ


It can be shown that the angle between two vectors x, y ∈ Rn satisfies

cos θ =x · y‖x‖ ‖y‖ where 0 ≤ θ ≤ π. (3.2.1)

Since, −1 ≤ cos θ ≤ 1, we can rewrite Equation 3.2.1 as

−1 ≤ x · y‖x‖ ‖y‖ ≤ 1. (3.2.2)

Remark 3.2.2

Let x, y ∈ Rn and 0 ≤ θ ≤ π, then

1. x ⊥ y (orthogonal or perpendicular) ⇐⇒ θ = π2⇐⇒ cos θ = 0⇐⇒ x · y = 0.

2. x // y (parallel in the same direction) ⇐⇒ θ = 0⇐⇒ cos θ = 1⇐⇒ x · y = ‖x‖ ‖y‖ ⇐⇒y = c x with c > 0.

3. x // y (parallel in the opposite direction) ⇐⇒ θ = π ⇐⇒ cos θ = −1 ⇐⇒ x · y =

−‖x‖ ‖y‖ ⇐⇒ y = c x with c < 0.

Example 3.2.3

Find all values of c so that x = (c, 2, 1, c) and y = (c,−1,−2,−3) are orthogonal.

Solution:

By Remark 3.2.2, we have x ⊥ y⇐⇒ x · y = 0. That is,

x · y = 0

c2 − 2− 2− 3c = 0

c2 − 3c− 4 = 0

(c− 4)(c + 1) = 0 c = 4 or c = −1.

Example 3.2.4

Find all vectors in R4 that are orthogonal to both

x = (1, 1, 1, 1) and y = (1, 1,−1,−1).


Solution:

Let z = (a, b, c, d) ∈ R4 so that z · x = z · y = 0. Therefore, we get the following homogenous

system:

a + b + c + d = 0

a + b− c− d = 0=⇒

1 1 1 1 0

1 1 −1 −1 0

∼

1 1 0 0 0

0 0 1 1 0

Solving this system, we get a = −b and c = −d. Let b = r and d = s where r, s ∈ R to get

z = (−r, r,−s, s) which is the form of any vector in R4 that is orthogonal to both x and y.

Example 3.2.5

Show that the triangle with vertices P1(2, 3,−4), P2(3, 1, 2), and P3(7, 0, 1) is a right triangle.

Solution:

P1 P2

P3

Figure 1

P1 P2

P3

−−−→P1P2

−−−→P1P3

−−−→P2P3

Figure 2

P1

P2P3

Figure 3

We start with Figure 1 as we do not know if there is a right angle. We create three vectors,

namely

x =−−→P1P2 = P2 − P1 = (1,−2, 6)

y =−−→P1P3 = P3 − P1 = (5,−3, 5)

z =−−→P2P3 = P3 − P2 = (4,−1,−1)

This is draw in Figure 2. Then, we want to find two vectors whose dot product is zero which is

valid by considering x and z. That is

x · z = 4 + 2− 6 = 0.

Therefore, this triangle has a right angle at P2 and it is drawn at Figure 3.


Example 3.2.6

Find the angle between x = (0, 1, 1, 0) and y = (1, 1, 0, 0).

Solution:

By Equation 3.2.1, we have cos θ = x·y‖x‖ ‖y‖

where x · y = 0 + 1 + 0 + 0 = 1 and

‖x‖ =√

02 + 12 + 12 + 02 =√

2 = ‖y‖.

Therefore, cos θ = 12

which implies that θ = π3.

Theorem 3.2.2: Cauchy-Schwarz Inequality

If x and y are in Rn, then |x · y| ≤ ‖x‖ ‖y‖.

Proof:

By Equation 3.2.1, we have cos θ = x·y‖x‖ ‖y‖

for 0 ≤ θ ≤ π. Thus,

∣∣∣∣∣

x · y‖x‖ ‖y‖

∣∣∣∣∣

= | cos θ| ≤ 1 where ‖x‖ ‖y‖ is always non-negative

|x · y|‖x‖ ‖y‖ ≤ 1 =⇒ |x · y| ≤ ‖x‖ ‖y‖.

Theorem 3.2.3: Triangle Inequality

If x and y are in Rn, then ‖x + y‖ ≤ ‖x‖+ ‖y‖.

Proof:

By Remark 3.2.1, we have

‖x + y‖2 = (x + y) · (x + y)

= x · x + x · y + y · x + y · y

= ‖x‖2 + 2x · y + ‖y‖2take absolute value for both sides.

‖x + y‖2 = ‖x‖2 + 2|x · y|+ ‖y‖2use Cauchy-Schwarz Inequality.

≤ ‖x‖2 + 2‖x‖‖y‖+ ‖y‖2


≤ (‖x‖+ ‖y‖)2take square root for both sides.

‖x + y‖ ≤ ‖x‖+ ‖y‖.

Remark 3.2.3

If x and y are in Rn, then ‖x− y‖ ≥

∣∣∣∣‖x‖ − ‖y‖

∣∣∣∣.

Proof:

We simply show that ‖x− y‖ ≥ ±(

‖x‖ − ‖y‖)

. First

‖x ‖ = ‖ (x− y) + y ‖ ≤ ‖x− y ‖+ ‖y ‖ → ‖x ‖ − ‖y ‖ ≤ ‖x− y ‖.

On the other hand, we have

− (‖x ‖ − ‖y ‖) = ‖y ‖ − ‖x ‖ ≤ ‖y− x ‖ = ‖x− y ‖.

TRUE or FALSE:

⋆ There are X, Y ∈ Rn with ‖X‖ = 2 and ‖Y ‖ = 4 and ‖X − Y ‖ = 1. (FALSE).

reason: By the previous remark: 1 = ‖X − Y ‖ ≥∣∣∣∣‖X‖ − ‖Y ‖

∣∣∣∣ =

∣∣∣∣2− 4

∣∣∣∣ = 2 which is not true.

Definition 3.2.3

The unit vector in Rn is a vector with length 1, i.e. u is a unit vector if ‖u‖ = 1.

Remark 3.2.4

Given a non-zero vector x ∈ Rn, u =

1

‖x‖x is a unit vector in the direction of x.

Example 3.2.7

Find a vector of length 3 in the opposite direction of x = (2, 2,−2).

Solution:


We first find a unit vector in the direction of x and then multiply it by ”−3” to get a vector of

length 3 in the opposite direction. That is,

u =1

‖x‖x =1√

4 + 4 + 4(2, 2,−2) =

1

2√

3(2, 2,−2) = (

1√3

,1√3

,−1√

3).

By Remark 3.2.4, u is a unit vector in the same direction of x. Therefore, y = −3u =

(−√

3,−√

3,√

3) is a vector of length 3 in the opposite direction of x.

Example 3.2.8

Show that if x and y are orthogonal unit vectors in Rn, then

‖a x + b y‖ =√

a2 + b2.

Solution:

By Remark 3.2.1, we have

‖a x + b y‖2 = (a x + b y) · (a x + b y)

= a2✟✟✟✯1‖x‖2 + 2 a b✘✘✘✘✿0

(x · y) + b2✟✟✟✯1‖y‖2

x · y = 0︸︷︷︸

orthogonal

and ‖x‖2 = ‖y‖2 = 1︸︷︷︸

unit vectors

= a2 + b2

Therefore, by taking the square root for both sides, we get ‖a x + b y‖ =√

a2 + b2.

3.2.5 Principle of Unit Vectors in Rn

⋆ Principle of Unit Vectors in R2:

Any vector x = (x1, x2) ∈ R2 can be written as

x = x1i + x2j, where i = (1, 0), j = (0, 1) ∈ R2.

O(0, 0)x

y

i

j


⋆ Principle of Unit Vectors in R3:

Any vector x = (x1, x2, x3) ∈ R3 can be written

as x = x1i + x2j + x3k, where i = (1, 0, 0), j =

(0, 1, 0), k = (0, 0, 1) ∈ R3.

O(0, 0, 0)

x

y

z

i

j

k

⋆ Principle of Unit Vectors in Rn:

Any vector x = (x1, x2, · · · , xn) ∈ Rn can be written as x = x1E1 + x2E2 + · · · + xnEn, where

E1 = (1, 0, · · · , 0), E2 = (0, 1, 0, · · · , 0), · · · , En = (0, · · · , 0, 1) ∈ Rn.


CHAPTER 3 REVISION


1. (From the book) Show that U · (V + W ) = U · V + U ·W , for any vectors U, V, W ∈ Rn.

2. (From the book) Show that ‖U + V ‖2 + ‖U − V ‖2 = 2‖U ‖2 + 2‖V ‖2, for any vectors

U, V ∈ Rn.

3. Show that if x and y are orthogonal unit vectors in Rn, then

‖ 4x + 3y ‖ = 5.

4. Let x, y ∈ Rn with ‖x ‖ = ‖y ‖. Show that x− y and x + y are orthogonal.

5. Let A =

cos (α) sin (α)

− sin (α) cos (α)

where α ∈ R. Show that if X ∈ R2, then ‖X ‖ = ‖AX ‖.

6. Let x, y ∈ Rn such that ‖x ‖ = 2 and ‖y ‖ = 3 and the angle between them is π

3. Evaluate

‖x− y ‖.7. Let x, y, and z be three unit vectors in R

n such that each vector is orthogonal to the other

two vectors. Show that ‖x + y + z ‖ =√

3.

8. If ‖x ‖ = 2 and ‖y ‖ = 3. What are the largest and the smallest possible values for

‖x− y ‖?9. Use Cauchy-Schwarz’s inequality to show that

(ab− cd + xy)2 ≤ (a2 + d2 + y2)(b2 + c2 + x2)

for all real numbers a, b, c, d, x, and y.

10. Find all vectors in R4 which are perpendicular to the vectors X = (1, 1, 2, 2) and Y =

(2, 3, 5, 5).

11. Let U, V ∈ Rn be unit vectors. Prove that (U + 2V ) · (2U − V ) ≤ 3.

12. Verify that the triangle with vertices A(1, 1, 2), B(1, 2, 3), and C(3, 0, 3) is a right triangle.

13. Let θ be the angle between the vectors U = (4,−2, 1, 2) and V = (4, 2, 5, 2). Find cos θ.

14. For any vectors X and Y in Rn, show that ‖X‖ ≤ ‖X − 2Y ‖+ 2‖Y ‖.

15. Let x and y be two vectors in Rn. Prove that ‖x− y‖ ≤ ‖x‖+ ‖y‖.

16. Find all values of a for which X = (a2 − a,−3,−1) and Y = (2, a− 1, 2a) are othogonal.


17. Find a vector X, of length 6, in the opposite direction of Y = (1, 2,−2).

18. Let X and Y be vectors in Rn such that ‖X‖ = ‖Y ‖. Show that X + Y and X − Y are

orthogonal vectors.

19. Find all x, y ∈ R such that the vectors U = (x, 1, 3) and V = (4, x + y, 9) are parallel.

[Hint: U and V are parallel iff there is r ∈ R such that U = rV .]

20. Let U and V be two vectors in R3 such that ‖U‖ = 2 and ‖V ‖ = 3.

(a) Find the maximum possible value for ‖2U + 3V ‖.(b) If U and V are orthogonal, find ‖2U + 3V ‖.

21. Let X, Y ∈ Rn. Find X · Y given that ‖X + Y ‖ = 1 and ‖X − Y ‖ = 5.

22. Answer each of the following as True or False:

(a) If U and V are two unit vectors in Rn, then ‖U − 6V ‖ ≥ 5.

(b) There exist X, Y ∈ R4 such that ‖X‖ = ‖Y ‖ = 2 and X · Y = 6.

5

Chapter

Real Vector Spaces

5.1 Real Vector Spaces

Definition 5.1.1

A real vector space V is a set of elements with two operations ⊕ and ⊙ satisfying the following

conditions. For short, we write (V,⊕,⊙) is a vector space if

(α) if x, y ∈ V, then x⊕ y ∈ V, that is ”V is closed under ⊕”: for all x, y, z ∈ V

(a) x⊕ y = y ⊕ x,

(b) x⊕ (y ⊕ z) = (x⊕ y)⊕ z,

(c) there exists 0 ∈ V such that x⊕ 0 = 0⊕ x = x,

(d) for each x ∈ V, there exists e ∈ V such that e⊕ x = x⊕ e = 0.

(β) if x ∈ V and c ∈ R, then c⊙ x ∈ V, that is ”V is closed under ⊙”: for all x, y ∈ V and for

all c, d ∈ R

(a) c⊙ (x⊕ y) = c⊙ x⊕ c⊙ y,

(b) (c + d)⊙ x = c⊙ x⊕ d⊙ x,

(c) c⊙ (d⊙ x) = (c d)⊙ x,

(d) 1⊙ x = x⊙ 1 = x.

Remark 5.1.1

(Rn, +, ·) is a vector space. That is, Rn with vector addition and scalar multiplication is a vector

space.

77

78 Chapter 5. Real Vector Spaces

Example 5.1.1

Consider V = {(x, y, z) : x, y, z ∈ R} with

(x1, y1, z1)⊕ (x2, y2, z2) = (x1 + x2, y1 + y2, z1 + z2),

and

c⊙ (x, y, z) = (cx, cy, 0).

Is (V,⊕,⊙) a vector space? Explain.

Solution:

Clearly, the α conditions are satisfied because this is the usual vector addition and hence V is

closed under ⊕. Thus, we only check on the β conditions. Let x = (x1, y1, z1), y = (x2, y2, z2)

be any two vectors in V, then

1. c ⊙ (x + y) = c ⊙ (x1 + x2, y1 + y2, z1 + z2) = (cx1 + cx2, cy1 + cy2, 0) = (cx1, cy1, 0) +

(cx2, cy2, 0) = c⊙ x + c⊙ y. This condition is satisfied.

2. (c + d) ⊙ x = ((c + d)x1, (c + d)y1, 0) = (cx1, cy1, 0) + (dx1, dy1, 0) = c ⊙ x + d ⊙ x. This

condition is satisfied.

3. c⊙ (d⊙ x) = c⊙ (dx1, dy1, 0) = (cdx1, cdy1, 0) = (cd)⊙ x. This condition is satisfied.

4. 1⊙ x = (x1, y1, 0) 6= (x1, y1, z1). This condition is NOT satisfied.

Therefore, (V,⊕,⊙) is not a vector space.

Example 5.1.2

Let V = {(x, y, z) : x, y, z ∈ R and z > 0} associated with the operations:

(x1, y1, z1)⊕ (x2, y2, z2) = (x1 + x2, y1 + y2, z1 + z2), and c⊙ (x, y, z) = (cx, cy, cz).

Is (V,⊕,⊙) a vector space? Explain.

Solution:

No. If c ∈ R with c < 0, then c⊙ (x, y, z) = (cx, cy, cz) 6∈ V since cz < 0.

5.1. Real Vector Spaces 79

Example 5.1.3

Is the set of real numbers under the substraction and scalar multiplication a vector space?

Explain.

Solution:

NO. Clearly, for x, y ∈ R, x⊕ y = x− y 6= y − x = y ⊕ x.


Exercise 5.1.1

1. Let V = {(x, y, 0) |x, y ∈ R}, be associated with the operations:

(x1, y1, 0)⊕ (x2, y2, 0) = (x1 + x2, y1 + y2, 0), and c⊙ (x, y, 0) = (cx, cy, 0).

Is (V,⊕,⊙) form a vector space? Explain. (This is a Vector Space!!).

2. Let V = {(x, y) | x, y ∈ R}. Define addition and scalar multiplication on V as follows: for

each (x, y), (x′, y′) ∈ V and a ∈ R,

(x, y)⊕ (x′, y′) = (x + x′, y + y′) and a⊙ (x, y) = (ay, ax).

Determine whether V with the given operations is a vector space. Justify your answer.

3. Consider R2 with the operations ⊕ and ⊙ where (x, y) ⊕ (x′, y′) = (2x − x′, 2y − y′) and

c⊙ (x, y) = c(x, y). Does the property (c + d)⊙X = c⊙X ⊕ d⊙X hold for all c, d ∈ R

and all X ∈ R2? Explain.

4. Consider the set V = {(x, y, z) : x, y, z ∈ R} with the following operations

(x1, y1, z1)⊕ (x2, y2, z2) = (x1 + x2, y1 + y2, z1 + z2) and c⊙ (x, y, z) = (z, y, x).

Is V a vector space? Explain.

5. Let V = {(x, y, z) : x, y, z ∈ R} and define

(x, y, z)⊕ (a, b, c) = (x + a, y + b, z + c) and k ⊙ (x, y, z) = (kx, ky, 0).

Show that V is not a vector space.

6. Consider the set V = {(x, y) |x, y ∈ R} with the following operations

(x, y)⊕ (x′, y′) = (x− x′, y − y′) and k ⊙ (x, y) = (kx, ky).

Determine whether (V,⊕,⊙) is a vector space (justify your answer).


(a) (T) V = {(x, y) ∈ R2 : y < 0} is closed under the operation c⊙ (x, y) = (cx, y).

(b) (F) V = {(x, y) ∈ R2 : y < 0} is closed under the operation c⊙ (x, y) = (cy, x).

5.2. Real Vector Subspaces 81

5.2 Real Vector Subspaces

Definition 5.2.1

Let (V,⊕,⊙) be a vector space and let W ⊆ V be non-empty. If (W,⊕,⊙) is a vector space,

then W is a subspace of V.

Remark 5.2.1

If V is a vector space, then V and {0} are subspaces of V. They are called trivial subspaces

of V.

Theorem 5.2.1

Let (V,⊕,⊙) be a vector space and let W be a subset of V. Then, W is a subspace of V if and

only if the following conditions hold:

1. W 6= φ,

2. for all x, y ∈W, x⊕ y ∈W,

3. for all x ∈W and c ∈ R, c⊙ x ∈W.

Example 5.2.1

Is W = {(x, y, 0, z2) : x, y ∈ R and z ∈ Z} a subspace of R4? Explain.

Solution:

1. (0, 0, 0, 0) ∈W and hence W is non-empty.

2. let (x1, y1, 0, z21), (x2, y2, 0, z2

2) ∈W, then

(x1, y1, 0, z21) + (x2, y2, 0, z2

2) = (x1 + x2, y1 + y2, 0, z21 + z2

2) 6∈W since z21 + z2

2 6= (z1 + z2)2.

For example, (0, 0, 0, 4), (0, 0, 0, 9) ∈W while the sum of them (0, 0, 0, 15) 6∈W. Therefore, W is

not a subspace of R4.


Example 5.2.2

Is W = {(x, y, z) : x + y + z = 1, where x, y, z ∈ R} a subspace of R3? Explain.

Solution:

Clearly, (0, 0, 0) 6∈W and hence W is not a vector space and it is not a subspace of R3.

Example 5.2.3

Let W = {(a, b, c, d) : d = 2a− b and c = a}. Is (W, +, ·) a subspace of R4? Explain.

Solution:

1. (0, 0, 0, 0) ∈W, then W 6= φ,

2. let x = (a1, b1, a1, 2a1 − b1) and y = (a2, b2, a2, 2a2 − b2). Then,

x + y = (a1, b1, a1, 2a1 − b1) + (a2, b2, a2, 2a2 − b2)

= (a1 + a2, b1 + b2, a1 + a2, 2(a1 + a2)− (b1 + b2)) ∈W.

3. for x = (a, b, a, 2a−b) ∈W and k ∈ R, we have k(a, b, a, 2a−b) = (ka, kb, ka, 2(ka),−kb) ∈W.

Therefore, W is a subspace of R4.

Example 5.2.4

Let x, y ∈ Rn and W = {z : z = ax + by, for a, b ∈ R}. Is W a subspace of Rn? Explain.

Solution:

1. z = 0x + 0y = O ∈W, implies that W 6= φ,

2. let z1 = ax + by and z2 = cx + dy be two vectors in W. Then,

z1 + z2 = (ax + by) + (cx + dy)

= (a + c)x + (b + d)y =⇒ ∈W,

3. for c ∈ R and z = ax + by ∈W, we have

cz = cax + cby ∈W for ca, cb ∈ R.


Therefore, W is a subspace of Rn.

Definition 5.2.2

Let Ax = O be a homogenous system for A ∈Mm×n and x ∈ Rn. We define the null space (or

the solution space of Ax = O) by

W = {x : Ax = O } ⊆ Rn,

and we show that W is a subspace of Rn in the following example.

Example 5.2.5

Let Ax = O for A ∈ Mm×n and x ∈ Rn. Show that (the null space) W = {x |Ax = O} is a

subspace of Rn.

Solution:

1. W 6= φ since Ax = O always has a solution (either trivial or non-trivial),

2. let x, y ∈W, then Ax = O and Ay = O. But, A(x + y) = Ax + Ay = O + O = O. Thus,

x + y ∈W,

3. for any c ∈ R and x ∈W, we have A(cx) = cAx = cO = O, thus cx ∈W.

Therefore, W is a subsapce of Rn.

Example 5.2.6

Let Ax = B with B 6= O and A ∈Mm×n and x ∈ Rn. Let W = {x : Ax = B} ⊆ R

n. Determine

whether W is a subspace of Rn.

Solution:

Clearly, W might be empty ”φ” if the system Ax = B has no solution. Thus, W is not a subspace

of Rn.


⋆ Linear Combination:

Definition 5.2.3

Let x1, x2, · · · , xn be vectors in a vector space V, a vector x ∈ V is called a linear combination

of the vectors x1, x2, · · · , xn if and only if x = c1x1 + c2x2 + · · · + cnxn for some real numbers

c1, c2, · · · , cn.

Example 5.2.7

Determine whether the vector x = (2, 1, 5) is a linear combination of the set of vectors {x1, x2, x3}where x1 = (1, 2, 1), x2 = (1, 0, 2), and x3 = (1, 1, 0).

Solution:

x is a linear combination of x1, x2, x3 if we find numbers c1, c2, c3 so that x = c1x1 + c2x2 + c3x3.

Consider c1(1, 2, 1) + c2(1, 0, 2) + c3(1, 1, 0) = (2, 1, 5), this is a system in three unknowns:

c1 + c2 + c3 = 2

2c1 + 0 + c3 = 1

c1 + 2c2 + 0 = 5

So, we solve the system:

1 1 1 2

2 0 1 1

1 2 0 5

r2−2r1→r2−−−−−−→r3−r1→r3

1 1 1 2

0 −2 −1 −3

0 1 −1 3

r2↔r3−−−→

1 1 1 2

0 1 −1 3

0 −2 −1 −3

r1−r2→r1−−−−−−→r3+2r2→r3

1 0 2 −1

0 1 −1 3

0 0 −3 3

− 13

r3→r3−−−−−→

1 0 2 −1

0 1 −1 3

0 0 1 −1

r1−2r3→r1−−−−−−→r2+r3→r2

1 0 0 1

0 1 0 2

0 0 1 −1

Therefore,

c1

c2

c3

=

1

2

−1

is a solution. Thus, x = x1 + 2x2 − x3 (check!).


⋆ The Span:

Definition 5.2.4

Let S = {x1, x2, · · · , xk} be a subset of a vector space V. Then, the set of all vectors in V that

are linear combination of the vectors in S is denoted by span S or span{x1, x2, · · · , xk}.

Example 5.2.8

Let S = {(1, 1, 0, 1), (1,−1, 0, 1), (0, 1, 2, 1)}. Determine whether

(a) x = (2, 3, 2, 3), and

(b) y = (0, 1, 2, 3)

belongs to span S.

Solution:

(a) Clearly, x belongs to span S only if x is a linear combination of S. Consider the system

c1(1, 1, 0, 1) + c2(1,−1, 0, 1) + c3(0, 1, 2, 1) = (2, 3, 2, 3).

1 1 0 2

1 −1 1 3

0 0 2 2

1 1 1 3

r2−r1→r2−−−−−−→r4−r1→r4

1 1 0 2

0 −2 1 1

0 0 2 2

0 0 1 1

r2−r4→r2−−−−−−→r3−2r4→r3

1 1 0 2

0 −2 0 0

0 0 0 0

0 0 1 1

− 12

r2→r2−−−−−→r3↔r4

1 1 0 2

0 1 0 0

0 0 1 1

0 0 0 0

r1−r2→r1−−−−−−→

1 0 0 2

0 1 0 0

0 0 1 1

0 0 0 0

Therefore, the system has a solution and x is a linear combination of S. Thus, x ∈ span S.

(b) We do the same thing for y by considering

c1(1, 1, 0, 1) + c2(1,−1, 0, 1) + c3(0, 1, 2, 1) = (0, 1, 2, 3).

1 1 0 0

1 −1 1 1

0 0 2 2

1 1 1 3

r2−r1→r2−−−−−−→r4−r1→r4

1 1 0 0

0 −2 1 1

0 0 2 2

0 0 1 3

r2−r4→r2−−−−−−→r3−2r4→r3

1 1 0 2

0 −2 0 −2

0 0 0 −4

0 0 1 3


The third row suggests that 0 = −4 which is impossible. Thus this system has no solution

and y is not linear combination of S. Thus, y 6∈ span S.

Theorem 5.2.2

Let S = {x1, x2, · · · , xk} be a set of vectors of a vector space V, then span S is a subspace of

V.

Proof:

Let W = span S = {z : z = c1x1 + c2x2 + · · ·+ ckxk} ⊆ V. Then,

1. W 6= φ since z = x1 + x2 + · · ·+ xk ∈W for ci = 1 for all i = 1, 2, · · · , k,

2. let z1 = c1x1 + c2x2 + · · ·+ ckxk and z2 = d1x1 + d2x2 + · · ·+ dkxk be two elements in W,

then

z1 + z2 = (c1 + d1)x1 + (c2 + d2)x2 + · · ·+ (ck + dk)xk ∈W,

3. for c ∈ R and z = c1x1 + c2x2 + · · ·+ ckxk ∈W, we have

z = c c1x1 + c c2x2 + · · ·+ c ckxk ∈W.

Therefore, W = span S is a subspace of V.

We now go back to show how can we solve Example 5.2.3 by three different method. One method

was already shown in that example.

Example 5.2.9

Let W = {(a, b, c, d) : d = 2a− b and c = a}. Is (W, +, ·) a subspace of R4? Explain.

Solution: Definition

We simply show it by the meaning of the definition of subspaces. Look at the Example 5.2.3.

Solution: Null Space

W = { (a, b, c, d) : 2a− b− d = 0 and a− c = 0 }.


That is W is the solution space of Ax = O where A =

2 −1 0 −1

1 0 −1 0

and x = (a, b, c, d).

Therefore, W is a subspace of R4.

Solution: The Span

W = { (a, b, a, 2a− b) : a, b ∈ R } = { a (1, 0, 1, 2) + b (0, 1, 0,−1) : a, b ∈ R }.

Therefore W = span{ (1, 0, 1, 2), (0, 1, 0,−1) }. That is W is a subspace of R4.


5.3 Linear Independence

Definition 5.3.1

Let S = {x1, x2, · · · , xn} be a set of vectors in a vector space V, if every vector in V is a linear

combination of S, then we say that S spans V or that V is spanned by S.

Example 5.3.1

Let S = {x1, x2, x3} where x1 = (1, 2, 1), x2 = (1, 0, 2), and x3 = (1, 1, 0). Does S spans R3?

Explain.

Solution:

Let x = (a, b, c) be any vector in R3. Consider

c1(1, 2, 1) + c2(1, 0, 2) + c3(1, 1, 0) = (a, b, c).

One way to decide if x is a linear combination of vectors of S is to solve this system as we have

seen in Example 5.2.7.

An easier way (since this is a square coefficient matrix) is by computing the determinant and

make sure that the∣∣∣ A

∣∣∣ 6= 0 where A is the matrix of coefficient. That is,

∣∣∣ A

∣∣∣ =

∣∣∣∣∣∣∣∣∣

1 1 1

2 0 1

1 2 0

∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣

2 0

1 2

∣∣∣∣∣∣

−∣∣∣∣∣∣

1 1

1 2

∣∣∣∣∣∣

= 4− 1 = 3 6= 0.

Since∣∣∣ A

∣∣∣ 6= 0, the system has a unique solution and x = (a, b, c) is a linear combination of

vectors in S. Therefore, S spans R3.

Example 5.3.2

Let S = {x1, x2} where x1 = (1, 1, 0), and x2 = (1, 1, 1). Does S spans R3? Explain.

Solution:

5.3. Linear Independence 89

Let x = (a, b, c) be any vector in R3. Consider

c1(1, 1, 0) + c2(1, 1, 1) = (a, b, c).

Note that we can not use the determinant argument here since we have no square matrix. Thus,

solving the system, we get

1 1 a

1 1 b

0 1 c

r2−r1→r2−−−−−−→

1 1 a

0 0 b− a

0 1 c

.

This system has no solution if b − a 6= 0. Therefore, x is not a linear combination of S and S

does not span R3.

Remark 5.3.1

The following statements can be shown easily:

1. the vectors i = (1, 0), j = (0, 1) spans R2,

2. the vectors i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) spans R3,

3. the vectors E1 = (1, 0, · · · , 0), E2 = (0, 1, 0, · · · , 0), · · · , En = (0, · · · , 0, 1) spans Rn.

Definition 5.3.2

The set of vectors S = {x1, x2, · · · , xn} in a vector space V is said to be linearly dependent

if there exist constants c1, c2, · · · , cn not all zeros, such that

c1x1 + c2x2 + · · ·+ cnxn = 0.

Otherwise, S is said to be linearly independent. That is, x1, x2, · · · , xn are linearly indepen-

dent if whenever, c1x1 + c2x2 + · · ·+ cnxn = 0, we must have c1 = c2 = · · · = cn = 0.


Example 5.3.3

Determine whether x1 = (1, 0, 1, 2), x2 = (0, 1, 1, 2), and x3 = (1, 1, 1, 3) in R4 are linearly

independent or linearly dependent? Explain.

Solution:

We solve the homogenous system: c1x1 + c2x2 + c3x3 = 0. That is,

1 0 1 0

0 1 1 0

1 1 1 0

2 2 3 0

r3−r1→r3−−−−−−→r4−2r1→r4

1 0 1 0

0 1 1 0

0 1 0 0

0 2 1 0

r3−r2→r3−−−−−−→r4−2r2→r4

1 0 1 0

0 1 1 0

0 0 −1 0

0 0 −1 0

−r3→r3−−−−→−r4→r4

1 0 1 0

0 1 1 0

0 0 1 0

0 0 1 0

r1−r3→r1−−−−−−→r2−r3→r2−−−−−−−→r4−r3→r4

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 0

Therefore, c1 = c2 = c3 = 0 and thus x1, x2, x3 are linearly independent.

Example 5.3.4

Determine whether the vectors x1 = (1, 2,−1), x2 = (1,−2, 1), x3 = (−3, 2,−1), and x4 =

(2, 0, 0) in R3 is linearly independent or linearly dependent? Explain.

Solution:

Consider the homogenous system: c1x1 + c2x2 + c3x3 + c4x4 = 0. This system has a non-trivial

solutions because the number of unknowns (4) is greater than the number of equations (3).

Therefore, x1, x2, x3, x4 are linearly dependent.

In addition, we can show that this set is linearly dependent by the mean of r.r.e.f. as follows:

1 1 −3 2 0

2 −2 2 0 0

−1 1 −1 0 0

→ · · · →

1 0 −1 1 0

0 1 −2 1 0

0 0 0 0 0

.

From the reduced system above, we see that (from the third column) x3 = −x1 − 2x2 and that

(from the fourth column) x4 = x1 + x2.


Remark 5.3.2

Let S = {x1, x2, · · · , xn} be a set of vectors in Rn and let A be an n× n matrix whose columns

are the n-vectors of S. Then,

1. if A is singular, then S is linearly dependent,

2. if A is non-singular, then S is linearly independent.

Example 5.3.5

For what values of α are the vectors (−1, 0,−1), (2, 1, 2), (1, 1, α) in R3 linearly dependent?

Explain.

Solution:

We want the vectors to be linearly dependent, so consider the system c1(−1, 0,−1)+c2(2, 1, 2)+

c3(1, 1, α) = (0, 0, 0). This system has non-trivial solutions only if∣∣∣ A

∣∣∣ = 0, where A is the

matrix whose columns are [−1, 0,−1]T , [2, 1, 2]T , and [1, 1, α]T . That is,

∣∣∣ A

∣∣∣ =

∣∣∣∣∣∣∣∣∣

-1 2 1

0 1 1

-1 2 α

∣∣∣∣∣∣∣∣∣

= 0 ⇐⇒ −(α− 2)− (2− 1) = 0 ⇐⇒ 2− α− 1 = 0 ⇐⇒ α = 1.

Therefore, if α = 1 the vectors are linearly dependent. Otherwise if α ∈ R\{1}, the vectors are

linearly independent.

Example 5.3.6

Suppose that S = {v1, v2, v3} is a linearly independent set of vectors in a vector space V. Show

that T = {w1, w2, w3} is also linearly independent set, where w1 = v1 + v2 + v3, w2 = v2 + v3,

and w3 = v3

Solution:


Consider the system c1w1 + c2w2 + c3w3 = 0. Therefore,

c1w1 + c2w2 + c3w3 = 0

c1(v1 + v2 + v3) + c2(v2 + v3) + c3(v3) = 0

c1v1 + c1v2 + c1v3 + c2v2 + c2v3 + c3v3 = 0

(c1)v1 + (c1 + c2)v2 + (c1 + c2 + c3)v3 = 0

But v1, v2, v3 are linearly independent, thus c1 = c1 + c2 = c1 + c2 + c3 = 0. Therefore,

c1 = c2 = c3 = 0 and hence T is linealry independent.

Example 5.3.7

Suppose that S = {v1, v2, v3} is a linearly dependent set of vectors in a vector space V. Show

that T = {w1, w2, w3} is also linearly dependent set, where w1 = v1, w2 = v1 + v2, and

w3 = v1 + v2 + v3

Solution:

Consider the system c1w1 + c2w2 + c3w3 = 0. Therefore,

c1w1 + c2w2 + c3w3 = 0

c1(v1) + c2(v1 + v2) + c3(v1 + v2 + v3) = 0

(c1 + c2 + c3)v1 + (c2 + c3)v2 + (c3)v3 = 0

But v1, v2, v3 are linearly dependent, thus at least one of c1 + c2 + c3, c2 + c3, and c3 is non-zero.

Therefore, one of c1, c2, c3 is non-zero and hence T is linealry dependent.

Theorem 5.3.1

If {x1, x2} is a linearly dependent set, then x1//x2.

Proof:

If {x1, x2} is linearly dependent, then c1x1 + c2x2 = 0 where c1 or c2 is non-zero. Assume that

c2 6= 0, then x2 = −c1

c2x1 which implies that x1//x2.


Theorem 5.3.2

Any subset of a vector space V contains the zero vector is a linearly dependent set.

Proof:

Assume that S = {x1, x2, · · · , xn} ⊆ V with xk = 0, then let ci = 0 for all i = 1, 2, · · · , n and

let ck = 1. Then, the system c1x1 + c2x2 + · · ·+ cnxn = 0 holds with ck is a non-zero. That is,

S is a linearly dependent set.

Theorem 5.3.3

If {x1, x2, · · · , xn} is a linearly dependent set, then we can express one of the vectors in terms

of the others.

Proof:

Consider the system c1x1 + c2x2 + · · ·+ cnxn = 0 with not all ci zeros, say ck 6= 0. Then, clearly

xk =(−c1

ck

)

x1 +(−c2

ck

)

x2 + · · ·+(−ck−1

ck

)

xk−1 +(−ck+1

ck

)

xk+1 + · · ·+(−cn

ck

)

xn.

Remark 5.3.3

The set of linearly independent vectors should be non-zeros distinct vectors.

Remark 5.3.4

Let S1, S2 be two subsets of a vector space V with S1 ⊆ S2. Then,

1. if S1 is linearly dependent, then S2 is linearly dependent,

2. if S2 is linearly independent, then S1 is linearly independent.


5.4 Basis and Dimension

Definition 5.4.1

A set S = {x1, x2, · · · , xn} of distinct non-zero vectors in a vector space V is said to be a basis

if and only if

1. S spans V (V = span S),

2. S is linearly independent set.

The dimension of V is the number of vectors in its basis and is denoted by dim(V).

Remark 5.4.1

The following sets are called standard basis:

1. {(1, 0), (0, 1)} ∈ R2,

2. {(1, 0, 0), (0, 1, 0), (0, 0, 1)} ∈ R3,

3. {E1 = (1, 0, · · · , 0), E2 = (0, 1, 0, · · · , 0), · · · , En = (0, · · · , 0, 1)} ∈ Rn.

Example 5.4.1

Show that the set S = {x1, x2, x3, x4} is a basis for R4, where

x1 = (1, 0, 1, 0), x2 = (0, 1,−1, 2), x3 = (0, 2, 2, 1), and x4 = (1, 0, 0, 1).

Solution:

To show that S is a basis for R4, we show that S is linearly independent and that S spans R

4.

1. S is linealy independent? Consider the homogenous system

c1(1, 0, 1, 0) + c2(0, 1,−1, 2) + c3(0, 2, 2, 1) + c4(1, 0, 0, 1) = (0, 0, 0, 0).

This system has a trivial solution if∣∣∣ A

∣∣∣ 6= 0, where A is the matrix of coefficients. That

5.4. Basis and Dimension 95

is,

|A| =

∣∣∣∣∣∣∣∣∣∣∣∣

1 0 0 1

0 1 2 0

1 −1 2 0

0 2 1 1

∣∣∣∣∣∣∣∣∣∣∣∣

c1−c4→c1=

∣∣∣∣∣∣∣∣∣∣∣∣

0 0 0 1

0 1 2 0

1 −1 2 0

−1 2 1 1

∣∣∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣∣

0 1 2

1 −1 2

−1 2 1

∣∣∣∣∣∣∣∣∣

= · · · = 1 6= 0.

Therefore, the system has trivial solution and thus S is linearly independent.

2. S spans R4? Let x = (a, b, c, d) ∈ R

4. Consider the non-homogenous system:

c1(1, 0, 1, 0) + c2(0, 1,−1, 2) + c3(0, 2, 2, 1) + c4(1, 0, 0, 1) = (a, b, c, d).

Since the∣∣∣ A

∣∣∣ 6= 0 where A is the matrix of coefficients, then this system has a unique

solution and thus S spans R4.

Therefore, S is a basis for R4.

Theorem 5.4.1

Let S = {x1, x2, · · · , xn} be a basis for a vector space V. Then, every vector in V can be written

in one and only one way as a linear combination of the vectors in S.

Proof:

Let x ∈ V. Since S is a basis of V, then S spans V. That is, we can write

x = c1x1 + c2x2 + · · ·+ cnxn, and (5.4.1)

x = d1x1 + d2x2 + · · ·+ dnxn. (5.4.2)

By substracting Equation 5.4.2 out of Equation 5.4.1, we get

0 = (c1 − d1)x1 + (c2 − d2)x2 + · · ·+ (cn − dn)xn.

But S is linearly independent set (it is a basis). Thus,

c1 − d1 = 0, c2 − d2 = 0, · · · , cn − dn = 0.

Therefore,

c1 = d1, c2 = d2, · · · , cn = dn


and hence x can be written in one and only one way as a linear combination of vectors in S.

Theorem 5.4.2

Let S = {x1, x2, · · · , xn} be a set of non-zero vectors of a vector space V and let W = span S.

Then, some subset of S is a basis for W.

5.4.2: ⋆ How to construct a basis?

Let V be a vector space and S = {x1, x2, · · · , xn} is a subset of V. The procedure to find a

subset of S that is a basis for W = span S is:

1. form the linear combination c1x1 + c2x2 + · · ·+ cnxn = 0,

2. form the augmented matrix of the homogenous system in step (1),

3. find the r.r.e.f. of the augmented matrix,

4. the vectors in S that are corresponding to columns which contains the leading 1’s form a

basis for W = span S.

Example 5.4.2

Let S = {x1, x2, x3, x4, x5} be a set of R4 where x1 = (1, 2,−2, 1), x2 = (−3, 0,−4, 3), x3 =

(2, 1, 1,−1), x4 = (−3, 3,−9, 6), and x5 = (9, 3, 7,−6). Find a subset of S that is a basis for

W = span S. Find dim(W).

Solution:

We consider the homogenous system: c1x1 + c2x2 + c3x3 + c4x4 + c5x5 = 0 to find a linearly

independent subset of S:

1 −3 2 −3 9 0

2 0 1 3 3 0

−2 −4 1 −9 7 0

1 3 −1 6 −6 0

≈ r.r.e.f.· · · · · · ≈

1 0 12

32

32

0

0 1 −12

32−5

20

0 0 0 0 0 0

0 0 0 0 0 0

The leading entries are pointing (appear) on the first two columns, namely columns 1 and

2. Therefore, {x1, x2} is linearly independent and it spans W. Thus, {x1, x2} is a basis for


W = span S and dim(W) = 2.

Remark 5.4.3

If W is a subspace of a vector space V, then dim(W) ≤ dim(V).

Theorem 5.4.3

If S = {x1, x2, · · · , xn} is a basis for a vector space V and T = {y1, y2, · · · , ym} is a linearly

independent set of V. Then m ≤ n

Theorem 5.4.4

If S = {x1, x2, · · · , xn} and T = {y1, y2, · · · , ym} are two bases for a vector space V, then

m = n.

Theorem 5.4.5

If S is a linearly independent set of vectors in a finite dimensional vector space V, then there is

a basis T for V which contains S.

Example 5.4.3

Find a basis for R4 that contains the vectors x1 = (1, 0, 1, 0) and x2 = (−1, 1,−1, 0).

Solution:

Consider the set S = {x1, x2, E1, E2, E3, E4}. The set S spans R4 but it contains some linearly

dependent vectors. In order to delete those, we follow the following procedure:

1 −1 1 0 0 0 0

0 1 0 1 0 0 0

1 −1 0 0 1 0 0

0 0 0 0 0 1 0

≈ r.r.e.f.· · · · · · ≈

1 0 0 1 1 0 0

0 1 0 1 0 0 0

0 0 1 0 −1 0 0

0 0 0 0 0 1 0

The leading entries pointing on the columns 1, 2, 3, and 6. Therefore, the set {x1, x2, E1, E4} is

a basis for R4 containing x1 and x2.


Theorem 5.4.6

Let V be an n -dimensional vector space and let S = {x1, x2, · · · , x n } be a set in V. Then,

1. if S is linearly independent, then it is a basis for V,

2. if S spans V, then it is a basis for V.

Example 5.4.4

Find all values of a for which S = {(a2, 0, 1), (0, a, 2), (1, 0, 1)} is a basis for R3.

Solution:

Since dim(R3) = 3 = size of S, it is enough to show that S is linearly independent (or it spans

R3) to show that it is a basis for R

3. Consider

c1(a2, 0, 1) + c2(0, a, 2) + c3(1, 0, 1) = (0, 0, 0).

Clearly, S is linearly independent if∣∣∣ A

∣∣∣ 6= 0, where A is the coefficient matrix. That is,

∣∣∣∣∣∣∣∣∣

a2 0 1

0 a 0

1 2 1

∣∣∣∣∣∣∣∣∣

= a(a2 − 1) 6= 0 =⇒ a 6= 0 and a 6= ±1.

Therefore, S is a basis for R3 if a ∈ R\{−1, 0, 1}.

Example 5.4.5

Find the dimension of the subspace of all vectors of the form (a, b, c, d) where c = a − b and

d = a + b (for a, b ∈ R).

Solution:

Let W = {(a, b, a− b, a + b) | a, b ∈ R} ⊆ R4. Let x be any vector of W, then

x = (a, b, a− b, a + b) = a(1, 0, 1, 1) + b(0, 1,−1, 1) ∈W.

Therefore, S = {(1, 0, 1, 1), (0, 1,−1, 1)} spans W.

Clearly, c1(1, 0, 1, 1) + c2(0, 1,−1, 1) = (0, 0, 0, 0) holds only if c1 = c2 = 0 which shows that S is


linearly independent. That is, S is a basis for W and dim(W) = 2.

Example 5.4.6

Find the dimension of the subspace of all vectors of the form (a + c, a + b + 2c, a + c, a− b) where

a, b, c ∈ R.

Solution:

Let W = {(a + c, a + b + 2c, a + c, a− b) | a, b, c ∈ R} ⊆ R4. Let x be any vector of W, then

x = (a + c, a + b + 2c, a + c, a− b) = a(1, 1, 1, 1) + b(0, 1, 0,−1) + c(1, 2, 1, 0) ∈W.

Therefore, S = {(1, 1, 1, 1), (0, 1, 0,−1), (1, 2, 1, 0)} spans W.

Consider c1(1, 1, 1, 1) + c2(0, 1, 0,−1) + c3(1, 2, 1, 0) = 0. Thus,

1 0 1

1 1 2

1 0 1

1 −1 0

≈ r.r.e.f.· · · · · · ≈

1 0 1

0 1 1

0 0 0

0 0 0

The leading entries pointing on the columns 1 and 2. Therefore, the set {(1, 1, 1, 1), (0, 1, 0,−1)}is a basis for W containing two vectors. Thus, the dimension of W is 2.

Example 5.4.7

Let S = {x1, x2, x3} be a basis for a vector space V. Show that T = {y1, y2, y3} is also a basis

for V, where y1 = x1 + x2 + x3, y2 = x2 + x3, and y3 = x3.

Solution:

Since S is a basis for V, then dim(V) = 3. But the size of T is also 3. Thus, it is enough to

show that T is linearly independent to show that it is a basis for V as well.

Consider the system c1y1 + c2y2 + c3y3 = 0. Therefore,

c1y1 + c2y2 + c3y3 = 0

c1(x1 + x2 + x3) + c2(x2 + x3) + c3(x3) = 0

(c1)x1 + (c1 + c2)x2 + (c1 + c2 + c3)x3 = 0


But x1, x2, x3 are linearly independent, thus c1 = c1 + c2 = c1 + c2 + c3 = 0. Therefore,

c1 = c2 = c3 = 0 and hence T is linealry independent.

5.5. Homogenous System 101

5.5 Homogenous System

Definition 5.5.1

The null space of A (the solution space of Ax = 0) is the set

{x : Ax = 0}.

Remark 5.5.1

To find a basis for the null space of A in Ax = 0, we do the following:

1. use the Guass-Jordan method to solve the homogenous system Ax = 0,

2. if the solutions of Ax = 0 is trivial, then the null space of A = {0}, when∣∣∣ A

∣∣∣ 6= 0, and

dim(null space of A) = 0,

3. if the solution of Ax = 0 is non-trivial with solutions x1, x2, · · · , xr, then any vector x in

the null space of A can be written as x = c1x1+c2x2+· · ·+crxr. Therefore, {x1, x2, · · · , xr}spans the null space of A, and it is a basis. The dimension of the null space of A is called

the nullity of A.

Example 5.5.1

Find a basis for the solution space of Ax = 0, where

A =

1 1 0 2

−2 −2 1 −5

1 1 −1 3

Solution:

We first solve the homogenous system:

1 1 0 2 0

−2 −2 1 −5 0

1 1 −1 3 0

r2+2r1→r2−−−−−−→r3−r1→r3

1 1 0 2 0

0 0 1 −1 0

0 0 −1 1 0

r3+r2→r3−−−−−−→

1 1 0 2 0

0 0 1 −1 0

0 0 0 0 0

The reduced system is

x1 + x2 + 2x4 = 0


x3 − x4 = 0

Let x4 = t and x2 = r for r, t ∈ R so that we get x3 = t and x1 = −r − 2t. Thus,

x =

−r − 2t

r

t

t

, where r, t ∈ R

Therefore,

x = r

−1

1

0

0

+ t

−2

0

1

1

.

Thus,{

[−1 1 0 0]T , [−2 0 1 1]T}

spans the null space of A and it is a linearly independent

set, so it is a basis for the null space of A and the nullity of A is 2.

Example 5.5.2

Find a basis for the solution space of Ax = 0, where

A =

1 0 0

1 4 −1

2 0 1

.

What is the nullity of A.

Solution:

We solve the homogenous system Ax = 0:

1 0 0 0

1 4 −1 0

2 0 1 0

r2−r1→r2−−−−−−→r3−2r1→r3

1 0 0 0

0 4 −1 0

0 0 1 0

r2+r3→r2−−−−−−→

1 0 0 0

0 4 0 0

0 0 1 0

14

r2→r2−−−−→

1 0 0 0

0 1 0 0

0 0 1 0

We have a trivial solution. Therefore, the null space of A is {0} and the nullity is 0.

5.6. The Rank of a Matrix and its Application 103

5.6 The Rank of a Matrix and its Application

Definition 5.6.1

Let A =

a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...


∈Mm×n.

The set of rows of A are:

x1 = (a11 a12 · · · a1n)

x2 = (a21 a22 · · · a2n)...

...

xm = (am1 am2 · · · amn)

∈ Rn

These rows span a subspace of Rn which is called the row space of A. Moreover, the row

rank of A = dim(row space of A).

Similarly, the columns of A are:

y1 =

a11

a21...

am1

, y2 =

a12

a22...

am2

, · · · , yn =

a1n

a2n...

amn

∈ Rm.

These columns span a subspace of Rm which is called the column space of A. Moreover, the

column rank of A = dim(column space of A).

Remark 5.6.1

Let A be any m× n matrix, then

1. the row rank of A = the column rank of A = the rank of A,

2. n = the nullity of A + the rank of A.


Theorem 5.6.1

If A and B are two m× n row equivalent matrices, then they have the same row space.

In Example 5.6.1, we illustrate how to find bases for the row and column spaces of a given matrix.

Example 5.6.1

Let A =

1 −2 0 3 −4

3 2 8 1 4

2 3 7 2 3

−1 2 0 4 −3

.

1. find a basis for the row space of A,

2. find a basis for the column space of A,

3. find a basis for the row space that contains only rows of A,

4. find a basis for the column space that contains only columns of A.

Solution:

1. To find a basis for the row space of A, we have to find the r.r.e.f. of A, then the set of

non-zero rows of the r.r.e.f. forms a basis for the row space.

1 −2 0 3 −4

3 2 8 1 4

2 3 7 2 3

−1 2 0 4 −3

≈ r.r.e.f.· · · · · · ≈

1 0 2 0 1 ←0 1 1 0 1 ←0 0 0 1 −1 ←0 0 0 0 0

Therefore, the set {(1, 0, 2, 0, 1), (0, 1, 1, 0, 1), (0, 0, 0, 1,−1)} forms a basis for the row space

of A. Note that the row rank of A = 3. (That is, nullity of A = 5 - 3 = 2).

2. To find a basis for the column space of A, we have to find a basis for the row space of AT .

Therefore,

AT =

1 3 2 −1

−2 2 3 2

0 8 7 0

3 1 2 4

−4 4 3 −3

≈ r.r.e.f.· · · · · · ≈

1 0 0 1124

←0 1 0 −49

24←

0 0 1 73←

0 0 0 00 0 0 0


Therefore, the set{

(1, 0, 0, 1124

), (0, 1, 0, −4924

), (0, 0, 1, 73)}

is a basis for the row space of AT

and it is a basis for the column space of A. The column rank of A = 3. (That is, nullity

of A = 5 - 3 = 2).

3. To find a basis for the row space of A that containsonly rows of A, we do as follows:

AT =

1 3 2 −1

−2 2 3 2

0 8 7 0

3 1 2 4

−4 4 3 −3

≈ r.r.e.f.· · · · · · ≈

1 0 0 1124

0 1 0 −4924

0 0 1 73

0 0 0 00 0 0 0↑ ↑ ↑

Then, the leading entries are pointing to column 1, column 2 , and column 3 in the r.r.e.f.

of AT which correspond to row 1, row 2, and row 3 in A. Thus,

{(1,−2, 0, 3,−4), (3, 2, 8, 1, 4), (2, 3, 7, 2, 3)}

forms a basis for the row space of A containing only rows of A.

4. To find a basis for the column space of A that only contains columns of A, we do the

following:

1 −2 0 3 −4

3 2 8 1 4

2 3 7 2 3

−1 2 0 4 −3

≈ r.r.e.f.· · · · · · ≈

1 0 2 0 1

0 1 1 0 10 0 0 1 −10 0 0 0 0↑ ↑ ↑

Then, the leading entries are pointing to column 1, column 2 , and column 3 in the r.r.e.f.

of A . Thus,

{(1, 3, 2,−1), (−2, 2, 3, 3), (3, 1, 2, 4)}

forms a basis for the column space of A containing only columns of A.


⋆ Rank and Singularity:

Theorem 5.6.2

An n× n matrix A is non-singular if and only if rank(A) = n.

Theorem 5.6.3

If A is an n× n matrix, then rank(A) = n if and only if∣∣∣ A

∣∣∣ 6= 0.

Theorem 5.6.4

Let A be an n×n matrix. The non-homogenous system Ax = B has a unique solution for every

n× 1 matrix B if and only if rank(A) = n.

Theorem 5.6.5

Let S = {x1, x2, · · · , xn} be a set of n vectors in Rn and let A be the matrix whose rows

(columns) are the vectors of S. Then, S is linearly independent if and only if∣∣∣ A

∣∣∣ 6= 0.

Theorem 5.6.6

Let A be an m × n matrix. The homogenous system Ax = 0 has a non-trivial solutions if and

only if rank(A) < n.

Example 5.6.2

Let A =

1 1 4 1 2

0 1 2 1 1

0 0 0 1 2

1 −1 0 0 2

2 1 6 0 1

. Find the rank of A and the nullity of A.

Solution:


To find the rank and the nullity of A, we find any basis of any kind of A. So,

1 1 4 1 2

0 1 2 1 1

0 0 0 1 2

1 −1 0 0 2

2 1 6 0 1

≈ r.r.e.f.· · · · · · ≈

1 0 2 0 1 ←0 1 2 0 −1 ←0 0 0 1 2 ←0 0 0 0 00 0 0 0 0

Therefore {(1, 0, 2, 0, 1), (0, 1, 2, 0,−1), (0, 0, 0, 1, 2)} is a basis for the row space of A and rank(A)

= 3 which implies that nullity of A = 5− 3 = 2.

Example 5.6.3

If A is a 3 × 7 matrix, what is the largest possible rank of A (or what is the largest possible

number of linearly independent columns of A)?

Solution:

The largest possible row rank of A is 3 while the largest possible column rank of A is 7, but row

rank of A = column rank of A. Therefore, the largest possible rank of A is 3. The same thing

applies for the largest possible number of linearly independent columns of A.

Example 5.6.4

If A is a 5× 3 matrix, show that A has linearly dependent rows.

Solution:

The largest possible rank of A is 3 and thus A must has at least two linearly dependent rows.

Example 5.6.5

Let A =

1 −2 0 3 −4

3 2 8 1 4

2 3 7 2 3

−1 2 0 4 −3

.


1. Find bases for the row and column spaces of A,

2. Find a basis for the null space of A.

3. Does x = (1, 2, 4, 3, 0) belong to the row space of A? Explain.

Solution:

1. To get bases for the row space and column spaces of A, we do the following:

1 −2 0 3 −4

3 2 8 1 4

2 3 7 2 3

−1 2 0 4 −3

≈ r.r.e.f.· · · · · · ≈

1 0 2 0 1 ←0 1 1 0 1 ←0 0 0 1 −1 ←0 0 0 0 0↑ ↑ ↑

On one hand, the set {(1, 0, 2, 0, 1), (0, 1, 1, 0, 1), (0, 0, 0, 1,−1)} forms a basis for the row

space of A. On the other hand, the set {(1, 3, 2,−1), (−2, 2, 3, 2), (3, 1, 2, 4)} forms a basis

for the column space of A that only contains columns of A, but this is fine since there is

no restrictions on the basis of column space of A mentioned in the question.

2. To find a basis for the null space of A, we use the r.r.e.f. of A found in part(1) to find the

reduced homogenous system:

x1 + 2x3 + x5 = 0

x2 + x3 + x5 = 0

x4 − x5 = 0

Let x5 = t, x3 = r, where t, r ∈ R to get

x =

−2r − t

−r − t

r

t

t

= r

−2

−1

1

0

0

+ t

−1

−1

0

1

1

.

Therefore, {(−2,−1, 1, 0, 0), (−1,−1, 0, 1, 1)} is a basis for the null space of A. The nullity

of A is 2 while the rank(A) = 3.

3. Yes. It is clear that x = (1)(1, 0, 2, 0, 1) + (2)(0, 1, 1, 0, 1) + (3)(0, 0, 0, 1,−1) where those

vectors are the vectors of the basis of the row space that were found in (1). It is also

possible to consider the non-homogenous system x = c1 (1, 0, 2, 0, 1) + c2 (0, 1, 1, 0, 1) +

c3 (0, 0, 0, 1,−1) to find the same answer.


CHAPTER 5 REVISION


1. Consider V = { (x, y) : xy = 0 } with the usual vector addition and scalar multiplication

in R2. Is (V, +, ·) a vector space? Explain.

6

Chapter

Diagonalization

6.1 Eigenvalues and Eigenvectors

Definition 6.1.1

Let A ∈Mn×n. The real number λ is called an eigenvalue of A if there exists a nonzero vector

x ∈ Rn so that

Ax = λx x 6= 0.

In this case, x is called an eigenvector associated with λ. This is called the eigenproblem.

Note that if A is non-singular, then A−1X = 1λX.

For example, let A =

1 −1

−1 1

and x =

1

−1

. Then,

Ax =

1 −1

−1 1

1

−1

=

2

−2

= 2

1

−1

= 2x.

Therefore, λ = 2 is an eigenvalue of A associated with eigenvector x.

Definition 6.1.2

Let A = (aij), 1 ≤ i, j ≤ n, then fA(λ) = |λIn − A| is called the characteristic polynomial.

⋆ Finding eigenvalues and eigenvectors of a given A ∈Mn×n

Remark 6.1.1

Ax = λx, x 6= 0,

λx− Ax = 0, x 6= 0,

(λIn − A)x = 0, x 6= 0,

111

112 Chapter 6. Diagonalization

This system has non-trivial solutions if (λIn − A) is singular. Therefore,

|λIn − A| = 0 ⇐⇒ fA(λ) = 0.

The eigenvalues are the roots of fA(λ).

Example 6.1.1

Let A =

1 0 2

1 0 0

0 0 −1

. Find the eigenvalues and the associated eigenvectors.

Solution:

We first compute the characteristic polynomial fA(λ) = |λI3 − A| = 0 as follows:∣∣∣∣∣∣∣∣∣

λ− 1 0 −2

−1 λ 0

0 0 λ + 1

∣∣∣∣∣∣∣∣∣

= λ

∣∣∣∣∣∣

λ− 1 −2

0 λ + 1

∣∣∣∣∣∣

= λ(λ− 1)(λ + 1) = 0.

which implies that

λ1 = −1 , λ2 = 0 , λ3 = 1 .

To find the associate eigenvectors, say x1, x2, x3, we solve the homogenous system

λi − 1 0 −2 0

−1 λi 0 0

0 0 λi + 1 0

(6.1.1)

for i = 1, 2, 3, as follows:

1. λ1 = −1 ⇒ (λ1I3 −A)x1 = 0, x1 = (a, b, c) 6= (0, 0, 0). Substitute λ1 = −1 in Equation

6.1.1 to get:

−2 0 −2 0

−1 −1 0 0

0 0 0 0

→ · · · →

1 0 1 0

0 −1 1 0

0 0 0 0

.

Thus,

a + c = 0, and − b + c = 0 =⇒ a = −c, and b = c.

6.1. Eigenvalues and Eigenvectors 113

Let c = t ∈ R\{0} to get

x1 =

−t

t

t

for t = 1, we get x1 =

−1

1

1

.

2. λ2 = 0 ⇒ (λ2I3 − A)x2 = 0, x2 = (a, b, c) 6= (0, 0, 0). Substitute λ2 = 0 in Equation

6.1.1 to get:

−1 0 −2 0

−1 0 0 0

0 0 1 0

→ · · · →

1 0 0 0

0 0 1 0

0 0 0 0

.

Thus, we get a = c = 0. Let b = t ∈ R\{0}. So,

x2 =

0

t

0


0

1

0

.


6.1.1 to get:

0 0 −2 0

−1 1 0 0

0 0 2 0

→ · · · →

1 −1 0 0

0 0 1 0

0 0 0 0

.

Thus, a− b = 0, and c = 0. If b = t ∈ R\{0}, then a = t as well. So,

x3 =

t

t

0


1

1

0

.

Example 6.1.2

Let A ∈Mn×n. Show that

1. the∣∣∣ A

∣∣∣ is the product of the eigenvalues of A,

2. the matrix is singular if and only if one of its eigenvalues is zero.

Solution:


1. Let λ1, λ2, · · · , λn be the eigenvalues of A. Thus,

fA(λ) = |λIn − A| = (λ− λ1)(λ− λ2) · · · (λ− λn),

setting λ = 0, fA(0) = |−A| = (−λ1)(−λ2) · · · (−λn),

(−1)n∣∣∣ A

∣∣∣ = (−1)n λ1λ2 · · ·λn,

∣∣∣ A

∣∣∣ = λ1λ2 · · ·λn.

2. (⇒) if A is singular, then∣∣∣ A

∣∣∣ = 0 = λ1λ2 · · ·λn. Therefore, λ1 = 0, λ2 = 0, · · · , or λn = 0.

Thus, A has at least one zero eigenvalue.

(⇐) if A has a zero eigenvalue, say λi where 1 ≤ i ≤ n, then

∣∣∣ A

∣∣∣ = λ1λ2 · · ·λi · · ·λn = 0.

Therefore, A is singular.

Example 6.1.3

Show that if A is an upper (lower) triangular matrix, then the eigenvalues of A are the diagonal

entries.

Solution:

Let A = (aij) where 1 ≤ i, j ≤ n be an upper triangular matrix, i.e. aij = 0 for all i > j. Recall

that the determinant of an upper triangular matrix is the product of its diagonal entries. Thus,

fA(λ) = |λIn − A| = (λ− a11)(λ− a22) · · · (λ− ann) = 0.

Therefore, λ1 = a11, λ2 = a22, · · · , λn = ann. Similar idea for the case of a lower triangular

matrix.

Example 6.1.4

Show that A and AT have the same eigenvalues.

Solution:

6.1. Eigenvalues and Eigenvectors 115

Let A be an n× n matrix. Clearly,

fA(λ) = |λIn − A| =∣∣∣(λIn − A)T

∣∣∣ =

∣∣∣λIT

n − AT∣∣∣ =

∣∣∣λIn − AT

∣∣∣ = fAT (λ).

The characteristic polynomial of A is the same as that of AT , thus A and AT have the same

eigenvalues.


6.2 Diagonalization

Definition 6.2.1

A matrix B is said to be similar to matrix A, denoted by B ≡ A, if there exists a non-singular

matrix P such that B = P −1AP.

⋆ Properties of Similar Matrices:

1. A ≡ A since A = I−1AI.

2. if B ≡ A, then A ≡ B.

Proof. If B ≡ A, then ∃P, P −1 such that B = P −1AP or PBP −1 = A. Let Q = P −1 to get

A = Q−1BQ. Thus, A ≡ B.

3. if A ≡ B and B ≡ C, then A ≡ C.

Proof.

A ≡ B ⇒ ∃ P, P −1 such that A = P −1BP,

B ≡ C ⇒ ∃ Q, Q−1 such that B = Q−1CQ.

Therefore,

A = P −1BP = P −1Q−1C QP = (QP )−1 C (QP ) ⇒ A ≡ C.

4. if A ≡ B, then∣∣∣ A

∣∣∣ =

∣∣∣ B

∣∣∣.

Proof. If A ≡ B, then there exists P, P −1 such that B = P −1 A P with |P | 6= 0. Therefore,

∣∣∣ B

∣∣∣ =

∣∣∣P −1 A P

∣∣∣ = |P −1|

∣∣∣ A

∣∣∣ |P | =

✓✓✓1

|P |∣∣∣ A

∣∣∣✚

✚|P | =∣∣∣ A

∣∣∣.

5. if A ≡ B, then AT ≡ BT .

6.2. Diagonalization 117

Proof. If A ≡ B, then there exist P, P −1 such that B = P −1 A P . Thus,

B = P −1 A P,

BT = (P −1 A P )T ,

BT = P T AT (P −1)T ,

BT = P T AT (P T )−1.

Let Q−1 = P T , to get BT = Q−1 AT Q. Therefore, BT ≡ AT .

Theorem 6.2.1

Similar matrices have the same eigenvalues.

Proof:

Let A and B be two similar n × n matrices. Then, there are P, P −1 such that B = P −1 A P .

Then,

fB(λ) = |λIn −B| =∣∣∣λIn − P −1 A P

∣∣∣

=∣∣∣P −1(λP P −1 − A)P

∣∣∣

= ✟✟✟|P −1| |λIn − A|✚✚|P |

= |λIn − A|

= fA(λ).

Therefore, A and B have the same set of eigenvalues since they have the same characteristic

polynomial.

Definition 6.2.2

A matrix A is diagonalizable if and only if A is similar to a diagonal matrix D, i.e.

D = P −1 A P with |P | 6= 0.

D: its diagonal entries are the eigenvalues of A,

P: its columns are the linearly independent eigenvectors of A.


Theorem 6.2.2

A matrix A has linearly independent eigenvectors if all of its eigenvalues are real and distinct.

Theorem 6.2.3

A matrix A is diagonalizable if and only if it has linearly independent eigenvectors.

6.2.1 Case 1: Diagonalization of a Matrix with Distinct Eigenvalues

Example 6.2.1

Let A =

1 0 2

1 0 0

0 0 −1

. If possible, find matrices P and D so that A is diagonalizable.

Solution:

Recall that in Example 6.1.1, we found that

λ1 = −1, λ2 = 0, λ3 = 1,

x1 =

−1

1

1

, x2 =

0

1

0

, x3 =

1

1

0

.

Since, we have real and distinct eigenvalues, the eigenvectors x1, x2, and x3 are linearly inde-

pendent. Thus, A is diagonalizable and

D =

−1 0 0

0 0 0

0 0 1

and P =

−1 0 1

1 1 1

1 0 0

.


6.2.2 Case 2: Diagonalization of a Matrix with Repeated Eigenvalues

Example 6.2.2

Find, if possible, matrices D and P so that D = P −1 A P , where

A =

0 0 1

0 1 2

0 0 1

.

Solution:

We first compute the characteristic polynomial fA(λ) = |λI3 − A| = 0 as follows:

∣∣∣∣∣∣∣∣∣

λ 0 −1

0 λ− 1 −2

0 0 λ− 1

∣∣∣∣∣∣∣∣∣

= 0

λ(λ− 1)(λ− 1) = 0

which implies that

λ1 = 1 , λ2 = 1 , λ3 = 0 .


λi 0 −1 0

0 λi − 1 −2 0

0 0 λi − 1 0

(6.2.1)


1. λ1 = λ2 = 1 ⇒ (λ1I3 − A)x1 = 0, x1 = (a, b, c) 6= (0, 0, 0). Substitute λ1 = λ2 = 1 in

Equation 6.2.1 to get:

1 0 −1 0

0 0 −2 0

0 0 0 0

.

Thus,

a− c = 0 ⇒ a = c,


−2c = 0.

Let b = t ∈ R\{0} to get

x1 =

0

t

0


0

1

0

At this point, we see that we have only one eigenvector associated to two eigenvalues. That is,

the eigenvectors are linearly dependent and there is no matrices P and D so that D = P −1 A P .

Thus, A is not diagonalizable.

Example 6.2.3

Find, if possible, matrices D and P so that D = P −1 A P , where

A =

1 0 0

0 0 0

−1 0 0

.

Solution:

We first compute the characteristic polynomial fA(λ) = |λI3 − A| = 0 as follows:

∣∣∣∣∣∣∣∣∣

λ− 1 0 0

0 λ 0

1 0 λ

∣∣∣∣∣∣∣∣∣

= 0

λλ(λ− 1) = 0

which implies that

λ1 = 0 , λ2 = 0 , λ3 = 1 .


λi − 1 0 0 0

0 λi 0 0

1 0 λi 0

(6.2.2)



1. λ1 = λ2 = 0 ⇒ (λ1I3 − A)x1 = 0, x1 = (a, b, c) 6= (0, 0, 0). Substitute λ1 = λ2 = 0 in

Equation 6.2.2 to get:

−1 0 0 0

0 0 0 0

1 0 0 0

.

Thus,

−a = 0,

a = 0.

Let b = t and c = r (not both zeros) be two real numbers to get

x1 =

0

t

r

= t

0

1

0

+ r

0

0

1

if

{

t = 1

r = 0

}

⇒ x1 =

0

1

0

, if

{

t = 0

r = 1

}

⇒ x2 =

0

0

1

.

Therefore, for λ1 = λ2 = 0, we got two linearly independent eigenvectors.


6.2.2 to get:

0 0 0 0

0 1 0 0

1 0 1 0

.

Thus,

b = 0,

a + c = 0 ⇒ a = −c.

Let c = t ∈ R\{0} to get

x3 =

t

0

−t


1

0

−1

Since, we have three linearly independent eigenvectors x1, x2, and x3, A is diagonalizable and

D =

1 0 0

0 0 0

0 0 0

and P =

1 0 0

0 1 0

−1 0 1

.


CHAPTER 6 REVISION


1. Let A =

1 −5 1

−1 −3 2

0 0 3

. Find the eigenvalues and the associated eigenvectors.

2. Let A =

1 0 2

0 2 0

3 0 2

.

(a) Show that (0, 1, 0) is an eigenvector of A.

(b) Determine all the eigenvalues of A.

(c) Find, if possible, a nonsingular matrix P and a diagonal matrix D such that D =

P −1AP .

3. Let A =

0 2 −1

1 −1 1

0 0 1

.

(a) Show that (−1, 1, 0) is an eigenvector of A.

(b) Find the eigenvalues of A.

(c) Find the rank of the matrix A + 2 I3.

(d) Is A diagonalizable? Explain.

4. Let A be a nonsingular matrix with an eigenvalue λ = −1. Show that λ = −1 is also an

eigenvalue of A−1.

5. Let A be a skew-symmetric 9× 9 matrix. Show that 0 is an eigenvalue of A.

6. (a) Prove that a square matrix A is singular if and only if it has an eigenvalue 0.

(b) Use part (a) to show that 0 is an eigenvalue of the matrix A =

2014 −2014 1001

2015 −2015 1002

2016 −2016 1003

.

7. Let A =

1 2 0

0 2 0

1 a 2

, where a ∈ R.

(a) Find all eigenvalues of A.

(b) For a = −2, determine whether A is diagonalizable.


(c) For a 6= −2, fin all eigenvectors of A.

8. Let A =

2 2 0

2 2 0

2 2 0

, where a ∈ R.

(a) Show that the eigenvalues of A are λ1 = λ2 = 0 and λ3 = 4.

(b) Determine all the eigenvectors of A.

(c) If possible, find a nonsingular matrix P and a diagonal matrix D such that P −1AP =

D.

9. Let A =

1 −1 −1 −1

0 0 −2 −1

0 −2 0 −3

0 0 0 1

, where a ∈ R.

(a) Find the eigenvalues of A.

(b) Is A diagonalizable? Explain.

(c) Find the eigenvalues of A−1.

10. (from the book) Let A and B be nonsingular n× n matrices. Show that AB−1 and B−1A

have the same eigenvalues.

11. (from the book) If A and B are nonsingular matrices, then show that AB and BA are

similar.

12. (from the book) Show that if A is diagonalizable, then:

(a) AT is diagonalizable.

(b) Ak is diagonalizable, for any positive integer k.

4

Chapter

Vectors in Rn

4.1 The Cross Product in Rn

Definition 4.1.1

Let x = x1i + x2j + x3k and y = y1i + y2j + y3k be two vectors in R3, then the cross product

of x and y, denoted by x× y, is defined by

x× y =

∣∣∣∣∣∣∣∣∣

i j k

x1 x2 x3

y1 y2 y3

∣∣∣∣∣∣∣∣∣

The cross product of x and y can be written as:

x× y =

∣∣∣∣∣∣

x2 x3

y2 y3

∣∣∣∣∣∣

i−∣∣∣∣∣∣

x1 x3

y1 y3

∣∣∣∣∣∣

j +

∣∣∣∣∣∣

x1 x2

y1 y2

∣∣∣∣∣∣

k

Note that x× y is orthogonal to both x and y. x

yx× y

Example 4.1.1

Find x× y where x = 2i + j + 2k and y = 3i− j− k.

Solution:

x× y =

∣∣∣∣∣∣∣∣∣

i j k

2 1 2

3 −1 −1

∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣

1 2

−1 −1

∣∣∣∣∣∣

i−∣∣∣∣∣∣

2 2

3 −1

∣∣∣∣∣∣

j +

∣∣∣∣∣∣

2 1

3 −1

∣∣∣∣∣∣

k = i + 8j− 5k.

125


Remark 4.1.1

It can be shown (Try it your self) that

i× i = 0 j× j = 0 k× k = 0

i× j = k j× k = i k× j = −i

i× k = −j j× i = −k k× i = jk j

i

+

+

+− −

−

⋆ Properties of cross product:

Let x, y, z ∈ R3 and c ∈ R. Then, (Can you prove the following propertes?)

1. x× y = −y× x,

2. x× (y + z) = x× y + x× z,

3. (x + y)× z = x× z + y× z,

4. c x× y = x× c y = c (x× y),

5. x× x = 0,

6. x× 0 = 0× x = 0,

7. x× (y× z) = (x · z)y− (x · y)z,

8. (x× y)× z = (x · z)y− (y · z)x.

Example 4.1.2

Again, find x× y where x = 2i + j + 2k and y = 3i− j− k.

Solution:

By Remark 4.1.1, we can get the same result as follows:

(2i + j + 2k)× (3i− j− k) = 0− 2k + 2j− 3k + 0− i + 6j + 2i + 0 = i + 8j− 5k.

4.1. The Cross Product in Rn 127

Example 4.1.3

(T #4

pg.263

)

Let x = x1i + x2j + x3k, y = y1i + y2j + y3k, and z = z1i + z2j + z3k be in R3. Show that

(x× y) · z =

∣∣∣∣∣∣∣∣∣

x1 x2 x3

y1 y2 y3

z1 z2 z3

∣∣∣∣∣∣∣∣∣

.

Solution:

L.H.S. = (x× y) · z =

∣∣∣∣∣∣∣∣∣

i j k

x1 x2 x3

y1 y2 y3

∣∣∣∣∣∣∣∣∣

· (z1i + z2j + z3k)

=

∣∣∣∣∣∣

x2 x3

y2 y3

∣∣∣∣∣∣

i−∣∣∣∣∣∣

x1 x3

y1 y3

∣∣∣∣∣∣

j +

∣∣∣∣∣∣

x1 x2

y1 y2

∣∣∣∣∣∣

k

· (z1i + z2j + z3k)

=

∣∣∣∣∣∣

x2 x3

y2 y3

∣∣∣∣∣∣

z1 −∣∣∣∣∣∣

x1 x3

y1 y3

∣∣∣∣∣∣

z2 +

∣∣∣∣∣∣

x1 x2

y1 y2

∣∣∣∣∣∣

z3 = R.H.S.

Remark 4.1.2

Let x, y, z ∈ R3. Then,

1. (x× y) · z = x · (y× z),

2. (x× y) · x = (x× y) · y = 0. That is, x× y is orthogonal to both x and y,

3. x · (y× x) = y · (y× x) = 0.

Example 4.1.4

Let x, y, z be in R3 such that (x × y) · z = 6. Find a) x · (y× z), b) 2x · (y× z), c) x · (z× y)

and d) x× (y× 4x).

Solution:

1. x · (y× z) = 6,

2. 2x · (y× z) = 12,


3. x · (z× y) = −6,

4. x · (y× 4x) = 0.

Example 4.1.5

Find a vector of length 12 so that it is perpendicular to both

x = 2i− j− 2k and y = 2i + j.

Solution:

The vector x × y is always orthogonal to both x and y. So, we compute that vector and make

its length equals to 12.

x× y =

∣∣∣∣∣∣∣∣∣∣∣

i j k

2 −1 −2

2 1 0

∣∣∣∣∣∣∣∣∣∣∣

= 2i− 4j + 4k and ‖x× y‖ =√

4 + 16 + 16 = 6.

Therefore, 16(x×y) is a unit vector and orthogonal to both x and y; while 12

(16(x×y)

)

= 2(x×y)

is a vector of length 12 and orthogonal to both x and y.

Theorem 4.1.1

Let x, y ∈ R3 have an angle θ between them. Then

‖x× y‖ = ‖x‖‖y‖ sin θ.

Proof:

By Equation 3.2.1, we have cos θ = x·y‖x‖‖y‖

. That is, ‖x‖2‖y‖2 cos2 θ = (x · y)2. Thus,

‖x× y‖2 = (x× y) · (x× y)

= x · [y× (x× y)]

= x · [(y · y)x− (y · x)y]

= (x · x)(y · y)− (x · y)(x · y)

= ‖x‖2‖y‖2 − (x · y)2

= ‖x‖2‖y‖2 − ‖x‖2‖y‖2 cos2 θ


= ‖x‖2‖y‖2(1− cos2 θ)

‖x× y‖2 = ‖x‖2‖y‖2 sin2 θ.

Note that sin θ is positive for 0 ≤ θ ≤ π. Taking the square root for both sides of the above

equation, we get

‖x× y‖ = ‖x‖‖y‖ sin θ.

⋆ Applications of the cross product:

Let x, y, z ∈ R3, then

1. x ⊥ y⇐⇒ θ = π2⇐⇒ sin θ = 1⇐⇒ ‖x× y‖ = ‖x‖‖y‖,

2. x//y⇐⇒ θ = 0 or π ⇐⇒ ‖x× y‖ = 0⇐⇒ x× y = 0,

3. Area of triangle:

A∆ =1

2ah

sin θ =h

‖x‖=⇒ h = ‖x‖ sin θ and if ‖y‖ = a,

A∆ =1

2‖y‖‖x‖ sin θ =

1

2‖x× y‖.

h

y

x

‖y‖ = a

‖x‖

θ

4. Area of parallel gram (two triangles):

A� = ‖x× y‖.

5. Volume of parallel piped:

V olume = |x · (y× z)| .

x

y

z


Example 4.1.6

Find the area of the triangle with vertices: P1(2, 2, 4), P2(−1, 0, 5), and P3(3, 4, 3).

Solution:

Let x =−−→P1P2 = P2 − P1 = (−3,−2, 1) and y =

−−→P1P3 = P3 − P1 = (1, 2,−1). Then,

x× y =

∣∣∣∣∣∣∣∣∣

i j k

−3 −2 1

1 2 −1

∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣

−2 1

2 −1

∣∣∣∣∣∣

i−∣∣∣∣∣∣

−3 1

1 −1

∣∣∣∣∣∣

j +

∣∣∣∣∣∣

−3 −2

1 2

∣∣∣∣∣∣

k = 0i− 2j− 4k.

Therefore, ‖x× y‖ =√

4 + 16 =√

20 = 2√

5 and A∆ = 12‖x× y‖ =

√5.

Example 4.1.7

Find the volume of the parallel piped with a vertex at the origin and edges x = i − 2j + 3k,

y = i + 3j + k, and z = 2i + j + 2k.

Solution:

Volume =∣∣∣x · (y× z)

∣∣∣ =

∣∣∣(x× y) · z

∣∣∣ =

∣∣∣∣∣∣∣∣∣

∣∣∣∣∣∣∣∣∣

1 −2 3

1 3 1

2 1 2

∣∣∣∣∣∣∣∣∣

∣∣∣∣∣∣∣∣∣

=∣∣∣1(6− 1)− 1(−4− 3) + 2(−2− 9)

∣∣∣ =

∣∣∣5 + 7− 22

∣∣∣ = | − 10| = 10.


Exercise 4.1.1

1. Show that two nonzero vectors X and Y in R3 are parallel, if and only if, X × Y = 0.

2. If U and V are nonzero vectors in R3 such that ‖(2U) × (2V )‖ = −4U · V , compute the

angle between U and V .

3. Find the area of the triangle whose vertices are P (1, 0,−1), Q(2,−1, 3) and R(0, 1,−2).

4. Let U and V be unit vectors in R3. Show that ‖U × V ‖2 + (U · V )2 = 1.

5. Let X and Y be two nonzero vectors in R3, with angle θ = π

3between them. Find ‖X×Y ‖,

if ‖X‖ = 3 and ‖ − 2Y ‖ = 4.

6. If X and Y are two vectors in R3, show that X × Y is orthogonal to X.


4.2 Planes and Lines in R3

Definition 4.2.1

A line in R3 is determined (specified) by a fixed point P0 = (x0, y0, z0) and a directional vector

U = (a, b, c). The line L through P0 and parallel to U consists of the points P (x, y, z) such that

(x, y, z) = (x0, y0, z0) + t(a, b, c), where t ∈ R. (4.2.1)

Such equation is written as

x = P0 + tU , where x = (x, y, z).

The parametric equation of line L (Equation

4.2.1):

x = x0 + at

y = y0 + bt

z = z0 + ct

t ∈ R

while the symmetric form of L is given by:

t :=x− x0

a=

y − y0

b=

z − z0

c.

x

y

z

U = (a, b, c)

L

P0(x0, y0, z0)

P (x, y, z)

x

O(0, 0, 0)

Example 4.2.1

Let P1(2,−2, 3), P2(−1, 0, 4), P3(−4, 2, 5) be three points in R3.

1. Find the parametric equation and the symmetric form of the line that passes through the

points P1 and P2.

2. Does P3 lies on the same line? Explain.

Solution:

1. Let U =−−→P1P2 = P2 − P1 = (−3, 2, 1) and let P0 = P1 be a fixed point on the line call it

4.2. Planes and Lines in R3 133

L. Then, the parametric equations of L are:

x = 2 − 3t

y = −2 + 2t

z = 3 + t

t ∈ R

while the symmetric form of L is:

x− 2

−3=

y − (−2)

2=

z − 3

1⇐⇒ 2− x

3=

y + 2

2= z − 3.

2. We have to check if P3 satisfies the parametric equation of L:

−4 = 2 − 3t → t = −6−3

= 2,

2 = −2 + 2t → t = 42

= 2,

5 = 3 + t → t = 5− 3 = 2.

Therefore, t = 2 and P3 lies on L. The same check can be done using symmetric form of

L.

Remark 4.2.1

Let U 1 = (a1, b1, c1) and U 2 = (a2, b2, c2) be two vectors associated with L1 and L2 so that

L1 :x− x1

a1

=y − y1

b1

=z − z1

c1

and L2 :x− x2

a2

=y − y2

b2

=z − z2

c2

. Then,

1. L1 ⊥ L2 ⇐⇒ U 1 ⊥ U 2 ⇐⇒ U 1 ·U 2 = 0,

2. L1//L2 ⇐⇒ U 1//U 2 ⇐⇒ U 1 ×U 2 = 0⇐⇒ U 2 = c U 1 for cR.

Example 4.2.2

Show that L1 : P1(4,−1, 4) and U 1 = (1, 1,−3) and L2 : P2(3,−1, 0) and U 2 = (2, 1, 1) intersect

orthogonally, and find the intersection point.

Solution:


Clearly, U 1 ·U 2 = 2 + 1− 3 = 0. Then, U 1 ⊥ U 2 =⇒ L1 ⊥ L2. To find the intersection point

P (x, y, z), we look for a point satisfying both parametric equations at the same time:

L1 : x = 4 + t1 , y = −1 + t1, and z = 4 − 3t1,

L2 : x = 3 + 2t2 , y = −1 + t2, and z = t2.

Clearly, since y = −1 + t1 = −1 + t2, we get t1 = t2. Substituting this in z = 4 − 3t1 = t2, we

get 4 − 3t1 = t1 which implies that t1 = 1 = t2. Therefore, the intersection point according to

L1 is P (4 + 1,−1 + 1, 4− 3), that is P (5, 0, 1).

Definition 4.2.2

The equation of a plane Π is determined by a fixed point P0(x0, y0, z0) contained in Π and a

normal directional vector N = (a, b, c) which is orthogonal to Π. A point P (x, y, z) lies on the

plane Π if and only if

N ⊥ −−→P0P ⇐⇒ N · −−→P0P = 0.

The general form of the equation of the plane

Π that passes through P0(x0, y0, z0) and its normal

vector is N = (a, b, c) is

a(x− x0) + b(y − y0) + c(z − z0) = 0.

Where the standard form of the plane Π is

ax + by + cz + d = 0.

L

P0

P

N

Remark 4.2.2

Assume that we want to find an equation of a plane Π containing three points P1(x1, y1, z1),

P2(x2, y2, z2), and P3(x3, y3, z3), then we can use either of the following ways:

1. For any point P (x, y, z) ∈ Π, we use the standard form of Π: ax + by + cz + d = 0 and

apply it for the points P1, P2, and P3. This is a homogenous system in a, b, c, and d. This


system has non-trivial solutions if

∣∣∣∣∣∣∣∣∣∣∣∣

x y z 1

x1 y1 z1 1

x2 y2 z2 1

x3 y3 z3 1

∣∣∣∣∣∣∣∣∣∣∣∣

= 0.

Solving this determinant, we get an equation in the standard form for Π.

2. Another way is to compute two contained vectors in Π namely x =−−→P1P2 and y =

−−→P1P3

and consider the normal vector to Π which is N = x × y = (a, b, c) which is orthogonal

to Π. Then, the general form of Π is a(x− x1) + b(y − y1) + c(z − z1) = 0.

Example 4.2.3

Let P1(2,−2, 1), P2(−1, 0, 3), P3(5,−3, 4), and P4(4,−3, 7) be four points in R3. Then,

1. Find an equation of the plane Π that passes through P1, P2, and P3.

2. Is P4 contained in Π? Explain.

Solution:

1. Let x =−−→P1P2 = P2 − P1 = (−3, 2, 2) and y =

−−→P1P3 = P3 − P1 = (3,−1, 3). These

two vectors are contained in Π while N = x × y = (8, 15,−3) is a normal vector to Π.

Therefore, a general form of Π is 8(x− 2) + 15(y + 2)− 3(z − 1) = 0. The standard form

of Π is

8x + 15y − 3z + 17 = 0.

2. P4 is contained in Π if it satisfies its equation:

8(4) + 15(−3)− 3(7) + 17 = −17 6= 0.

Therefore, P4 is not on the plane Π.


Remark 4.2.3

Let Π1 : a1x + b1y + c1z + d1 = 0 and Π2 : a2x + b2y + c2z + d2 = 0. Then,

1. Π1//Π2 ⇐⇒N 1//N 2 ⇐⇒N 1 ×N 2 = 0⇐⇒N 2 = c N 1, where c ∈ R,

2. Π1 ⊥ Π2 ⇐⇒N 1 ⊥N 2 ⇐⇒N 1 ·N 2 = 0.

N1

N2

Π1

Π2

Π1//Π2

N1

N2

Π1 ⊥ Π2

Example 4.2.4

Find the parametric equation of the intersection line of the two planes:

Π1 : x− y + 2z = 3 and Π2 : 2x + 4y − 2z = −6.

Solution:

We form a non-homogenous system to solve for the parametric equation of the intersection line:

1 −1 2 3

2 4 −2 −6

r2−2r1→r2−−−−−−→

1 −1 2 3

0 6 −6 −12

16

r2→r2−−−−→

1 −1 2 3

0 1 −1 −2

r1+r2→r1−−−−−−→

1 0 1 1

0 1 −1 −2

Therefore, the reduced system is:

x + z = 1 and y − z = −2.

Let z = t ∈ R to get the parametric equation of the intersection line:

x

y

z

=

1− t

−2 + t

t


Example 4.2.5

Find two equations of two planes whose intersection line is the line L:

x = −2 + 3t; y = 3− 2t; z = 5 + 4t; where t ∈ R.

Solution:

The symmetric form of L is:x + 2

3=

y − 3

−2=

z − 5

4.

Therefore, a first plane is by equating x+23

= y−3−2

, to get−2x−4 = 3y−9. Thus, Π1 : 2x+3y−5 =

0. Another plane is by equating x+23

= z−54

, to get 4x + 8 = 3z−15. Thus Π2 : 4x−3z + 23 = 0.

Remark 4.2.4

Let U = (a1, b1, c1) be associated with the line L : x−x0

a1= y−y0

b1= z−z0

c1and N = (a2, b2, c2) be

associated with the plane Π2 : a2x + b2y + c2z + d2 = 0. Then,

1. L ⊥ Π⇐⇒ U//N ⇐⇒ U ×N = 0⇐⇒ U = c N where c ∈ R,

2. L//Π⇐⇒ U ⊥N ⇐⇒ U ·N = 0.

N1

Π

L ⊥ Π

L

U

N1

U

L

L//Π

Π

Example 4.2.6

Find a plane that passes through the point (2, 4,−3) and is parallel to the plane −2x + 4y −5z + 6 = 0.

Solution:


Since the two planes parallel, we can choose the normal vector of the second plane. That is

N = (−2, 4,−5). Thus, the equation of the plane is

−2(x− 2) + 4(y − 4)− 5(z + 3) = 0 =⇒ −2x + 4y − 5z − 27 = 0.

Example 4.2.7

Find a line that passes through the point (−2, 5,−3) and is perpendicular to the plane 2x −3y + 4z + 7 = 0.

Solution:

The line L is perpendicular to our plane. So, it is parallel to its normal vector, so we can choose

the normal vector as U . That is U = (2,−3, 4) and hence the parametric equation of L is

x = −2 + 2t

y = 5 − 3t

z = −3 + 4t

t ∈ R

Example 4.2.8

Show that the plane Π : 6x − 4y + 2z = 0 and the line L : x−13

= −y+42

= z − 5 intersect

orthogonally. Find the intersection point.

Solution:

We first have to write the symmetric form as

L :x− 1

3=

y − 4

−2=

z − 5

1.

Then, the normal vector of Π is N = (6,−4, 2) and the directional vector of L is U = (3,−2, 1).

Clearly, N = 2U which implies that N//U ⇐⇒ Π ⊥ L.

The intersection point with respect to L is

x = 1 + 3t

y = 4 − 2t

z = 5 + t

t ∈ R


Therefore, plugin these values into the plane equation, we get

6(1 + 3t)− 4(4− 2t) + 2(5 + t) = 0

6 + 18t− 16 + 8t + 10 + 2t = 0

28t = 0

Therefore, we get t = 0. Substituting this in the parametric equatio, we get the intersection

point as P (1, 4, 5).


Exercise 4.2.1

1. Consider the planes:

Π1 : x + y + z = 3, Π2 : x + 2y − 2z = 2k, and Π3 : x + k2z = 2.

Find all values of k for which the intersection of the three planes is a line.

[Hint: Any point on the intersection of the three planes must satisfies the three equa-

tions. This would give a system of three equation. This system must have infinitely many

solutions to describe a line.]

2. Consider the lines:

L1 :x + 1

3=

y + 4

2= z − 1, and L2 :

x− 3

2=

y − 4

−4=

z − 2

2.

(a) Show that L1 and L2 are perpendicular and find their point of intersection.

(b) Find an equation of the plane Π that contains both L1 and L2.

[Hint: (a) Show that U 1 ·U 2 = 0 and then find a point satisfying both equations of x, y,

and z in terms of t1 and t2, for instance. (b) Consider N = U 1 ×U 2.]

3. Let L be the line through the points P1(−4, 2,−6) and P2(1, 4, 3).

(a) Find parametric equations for L.

(b) Find two planes whose intersection is L.

4. Find the parametric equations for the line L which passes through the points P (2,−1, 4)

and Q(4, 4,−2). For what value of k is the point R(k + 2, 14,−14) on the line L?

5. Find the point of intersection of the line x = 1 − t, y = 1 + t, z = t, and the plane

3x + y + 3z − 1 = 0.

6. Find the equations in symmetric form of line of intersection of planes:

Π1 : x + 2y − z = 2, and Π2 : 3x + 7y + z = 11.

7. Find an equation of the plane containing the lines

L1 : x = 3 + t, y = 1− t, z = 3t, and L2 : x = 2s, y = −2 + s, z = 5− s.

8. Find a, b ∈ R so that the point P (3, a− 2b, 2a + b) lies on the line

L : x = 1 + 2t, y = 2− t, z = 4 + 3t.


CHAPTER 4 REVISION


1. Consider the lines:

L1 :x− 7

2=

y + 3

−1= z − 1, and L2 : x + 1 =

y + 6

3=

z − 2

−2.

(a) Show that L1 and L2 intersect and find their point of intersection.

(b) Find parametric equations of L3 which is perpendicular to both L1 and L2 and passes

through their point of intersection.

[Hint: (a) Show that U 1 × U 2 6= 0. Point of intersection is P (1, 0,−2). (b) Consider

U 3 = U 1 ×U 2 which is orthogonal to both U 1 and U 2.]

2. Find an equation of the plane Π through the point P (0, 1, 1) and containing the line

x = 1 + t, y = 1 + 2t, and z = −2 + 3t. [Hint: Consider the points Q(1, 1,−2) and

R(2, 3, 1) which are on the line. Then a normal vector of Π is−→PQ×−→PR.]

3. Find an equation of the plane Π passing throught the point (6, 0, 2) and is perpendicular

to the line L of intersection of the planes

Π1 : x + y + 2z = 4 and Π2 : 3x− 2y + z = 1.

[Hint: Consider N = N 1 ×N 2 = (1, 1, 2)× (3,−2, 1) as a normal vector to Π.]

4. Let L1 be the line of intersection of the planes:

x + 2y − z − 4 = 0 and 2x + 5y − 3z − 9 = 0.

(a) Find parametric equations for L1.

(b) Show that L1 and L2 : x = 2 + 3t, y = 1 + 2t, z = 4 + t, are perpendicular lines, but

they do not intersect.

(c) Find an equation for the plane that contains L1 and it is parallel to L2.

5. (a) Find parametric equations for the line L1 which passes through the points P (1, 2, 3)

and Q(4, 3, 2).

(b) Does L1 intersect the line L2 : x−12

= y−31

= z−23

? If so, find their point of intersection.


6. (a) Find an equation of the plane Π through the points P (2, 1,−2), Q(−1, 1, 0), and

R(3,−1,−1).

(b) Find all values of k for which the point (1, 1− 2k, 3k − 1) lies on the plane Π.

7. Let P (2,−2, 1), Q(−1, 0, 3), R(5,−3, 2), and S(7, 17,−6) be points in R3.

(a) Find an equation of the plane Π through P , Q, and R and show that S does not lie

on Π.

(b) Find the point T on the plane Π such that the vector−→ST is orthogonal to Π. [Hint:

Find parametric equation for the line through S and perpendicular to Π.]

8. Determine if the following parametric equations represent the same line

L1 : x = 4− t, y = 2 + 3t, z = 1 + 2t, and L2 : x = 1 + 3r, y = 11− 9r, z = 7− 6r.

9. Let L be the line which is perpendicular to the plane Π : 3x − y + 4z = 0, and passes

through the point P (2, 6, 7). Find symmetric equations for L.

10. Let Π1 : 2x + 8y + 5z = 3 and Π2 : x + 4y + 2z = 1 be two planes, and let L be the line of

intersection of Π1 and Π2. Find parametric equations for L.

11. Write an equation for the plane which contains the two lines:

L1 :x− 1

1=

y + 1

2=

z

−3and L2 : x = 1− t, y = 3− 2t, z = 2 + 3t.

12. Let A(1, 1, 1), B(1, 2, 2) and C(2, 3, 4) be three points in R3.

(a) Find the area of the triangle whose vertices are A, B and C.

(b) Find parametric equations of the line L passing through the points B and C.

(c) Find an equation of the plane Π containing the point A and orthogonal to−−→BC.

(d) Find the point of intersection, if any, between the line L and the plane Π.


(a) (T) The plane 5y + 3z − 7 = 0 is parallel to the vector (1, 0, 0).

(b) (T) In R3,if X = 4Y , then X × Y = (0, 0, 0).

(c) (F) In R3, the plane π : y = 3 is parallel to the y-axis.

(d) (T) The vector X × Y is perpendicular to 2X + 3Y .

Index

addition

closed, 66

additive

identity, 14

inverse, 14

adjoint, 56

basis, 94

standard, 94

Cauchy-Schwarz Inequality, 71

characteristic polynomial, 111

cofactor, 53

cofactor expansion, 53

column rank, 103

column space, 103

cross product, 125

determinant, 43

diagonalizable, 117

diagonalization, 111, 116

dimension, 94

dot product, 6, 66

eigenvalue, 111

eigenvector, 111

elementary row operations, 21

identity

additive, 14

matrix, 3

inner product, 6

inverse

additive, 14

leading entry, 20

line, 132

parametric equation, 132

linear combination, 8, 84

linear independence, 88

linearly dependent, 89

linearly independent, 89

matrix, 1

addition, 4

adjoint, 56

cofactor, 53

diagonal, 3

identity, 3

inverse, 31

invertable, 31

non-singular, 31

scalar, 3

similar, 116

singular, 31

skew-symmetric, 16

square, 1

substraction, 4

symmetric, 16

transpose, 4

triangular

lower, 3

upper, 3

143

144 INDEX

unit, 3

zero, 2

minor, 53

null space, 83, 101

nullity, 101

parallel gram, 129

parallel piped, 129

permutation, 43

even, 43

inversion, 43

odd, 43

sign, 43

plane, 132, 134

general form, 134

standard form, 134

rank, 103, 106

column, 103

row, 103

real vector space, 77

rearrangement, 43

reduced row echelon form, 20

row equivalent, 21

row rank, 103

row space, 103

scalar maltiplication, 5

scalar multiplication

closed, 66

similar, 116

singularity, 106

solution

infinite, 22, 24

no, 22, 25

non-trivial, 22, 27

trivial, 22, 26

unique, 22, 23

space

null, 101

solution of Ax = 0, 101

span, 85

spanned by, 88

spans, 88

standard form, 134

subspace, 81

trivial, 81

system

consistent, 27

homogenous, 22, 26, 101

inconsistent, 27

non-homogenous, 22, 23

triangle, 129

Triangle Inequality, 71

vector, 1, 65, 125

angle, 68

cross product, 125

directional, 132

distance, 67

dot product, 66

head, 65

length, 67

INDEX 145

norm, 67

normal, 134

orthogonal, 69

parallel, 69

perpendicular, 69

space, 77

tail, 65

unit, 72, 73

lecture notes in linear algebralecture notes in linear algebra dr. abdullah al-azemi mathematics...

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