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Homework Solutions for Math*1160 (W18).
MARCUS R. GARVIE ∗
July 4, 2017
∗Department of Mathematics & Statistics, University of Guelph
STOP!
Before looking at the answers the best study strategy is to:
1. First read your lecture notes for the relevant section.
2. Then attempt the questions without first looking at the answers.
3. Finally, if you are stuck, look at the general approach in the relevant answer andtry again.
Remember, struggle is usually necessary for effective learning.
NoteWe use the following abreviations:
d.p. = decimal places, e.g. π to 3 d.p. is 3.141s.f. = significant figures, e.g. π to 3 s.f. is 3.14
1
Contents
1 Linear Equations & Matrices 31.1 Formulation of systems of linear equations . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Algebraic Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . 81.6 Special Types of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2 Solving Linear Systems 162.1 Echelon Form of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Finding A−1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3 Determinants 253.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Properties of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3 Cofactor Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4 Inverse of a Matrix (via the Adjoint) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4 Real Vector Spaces 324.1 Vectors in the Plane and in 3-Space (generalized to n-Space) . . . . . . . . . . . . . . . . 324.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.4 Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.5 Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.6 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.7 Homogeneous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.8 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
5 Inner Product Spaces 595.1 Length and Direction in Rn (n ≥ 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.2 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
6 Eigenvalues and Eigenvectors 656.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656.2 Diagonalization and Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
2
Chapter 1
Linear Equations & Matrices
1.1 Formulation of systems of linear equations
1) A bartender makes a cocktail with whisky and rum that contains 10 ml of alcohol and 40 ml of water.The rum has 14% alcohol by volume (and hence 86% water by volume), while the whisky has 43%alcohol by volume (and hence 57% water by volume). Write 2 equations in 2 unknowns for the amountof whisky (x) and the amount of rum (y) needed. (Hint: see the similar problem in your lecture notes.)
We formulate two equation for the total alcohol and the total water content of the drink:
Alcohol:43
100x +
14
100y = 10
Water:57
100x +
86
100y = 40
I.e., {0.43x + 0.14y = 10
0.57x + 0.86y = 40
(x = 10.3 ml (1d.p.), y = 39.7 ml (1 d.p.))
1.2 Systems of Linear Equations
2) Apply back-substitution to solve the following linear system:
2x1 − x2 + 3x3 − 2x4 = 1
x2 − 2x3 + 3x4 = 2
4x3 + 3x4 = 3
4x4 = 4
3
Solution:Starting from the last equation:
4x4 = 4 =⇒ x4 = 1.
4x3 + 3(1) = 3 =⇒ x3 = 0.
x2 − 2(0) + 3(1) = 2 =⇒ x2 = −1.2x1 − (−1) + 3(0)− 2(1) = 1 =⇒ x1 = 1.
Thus the solution is (1,−1, 0, 1)
3) Find the unique solution of the following linear system:x1 + 2x2 + x3 = 3
3x1 − x2 − 3x3 = −12x1 + 3x2 + x3 = 4
Solution:As in the class examples we denote the ith equation by Ei. Do the following:
replace E2 with E2 − 3E1, then replace E3 with E3 − 2E1
yielding: x1 + 2x2 + x3 = 3
− 7x2 − 6x3 = −10− x2 − x3 = −2
Then:replace E3 with E3 − (1/7)E2, yielding:
x1 + 2x2 + x3 = 3
− 7x2 − 6x3 = −10− 1
7x3 = − 4
7
Back-substitution yields the solutionx3 = 4, x2 = −2, x1 = 3.
4) By Attempting to solve the following system show why no solution exists:{2x + 3y = 6
4x + 6y = 9
Solution:Replace E2 with E2 − 2E1 yielding {
2x + 3y = 6
0 = −3
The 2nd equation is impossible, thus the system is inconsistent.
5) Find the infinite solution set corresponding to the following linear system:
4
{2x + 3y = 6
4x + 6y = 12
Solution:Replace E2 with E2 − 2E1 yielding {
2x + 3y = 6
0 + 0 = 0
The 2nd equation is redundant (yields no information). Thus from the first equation we have
2x = 6− 3y
or x = 3−3
2y,
where y is free to be chosen (a ‘free variable’). E.g., if y = 2/3 then x = 3 − 1 = 2. In general, ify = α ∈ R then x = 3− (3/2)α. Thus there are an infinite number of solutions belonging to the set:{(
3−3
2α,α
)|α ∈ R
}.
6) (a) Without doing any row operations, explain why the following system of linear equations is consistent:2x1 + 3x2 + 5x3 = 0
−5x1 + 6x2 − 17x3 = 0
7x1 − 4x2 + 3x3 = 0
(b) Without doing any row operations, explain why the following system of linear equations has aninfinite number of solutions?
2x1 + 3x2 + 5x3 + 2x4 = 0
−5x1 + 6x2 − 17x3 − 3x4 = 0
7x1 − 4x2 + 3x3 + 13x4 = 0
Solution:(a) Homogeneous linear systems of the form Ax = 0 always possess the zero solution, i.e. here it isobvious that a solution is x1 = x2 = x3 = 0.
(b) Homogeneous linear systems either have only the unique zero (‘trivial’) solution, or have an infinitenumber of solutions. Here we have three equations in four unknowns. This means after applying rowoperations to get the associated augmented matrix in row echelon form we will have one free variable,leading to an infinite number of solutions.
Note:For those of you who want more details we can use techniques from Section 2.2∗ to solve this system.
∗Of course if you haven’t covered that section yet you can still apply the ‘method of elimination’ (page 21) to get the sameanswer, but it’s going to be a lengthy calculation.
5
I reduced the coefficient matrix to RREF using elementary row operations to get†1 0 0 533/2610 1 0 2/2610 0 1 −110/261
So the associated linear system is
x1 + 533261x4 = 0
x2 + 2261x4 = 0
x3 − 110261x4 = 0
So we see that x4 is a free variable, leading to the infinite solution set (exercise){(−
533
261α,−
2
261α,
110
261α,α) | α ∈ R
}.
E.g., taking α = 261 yields the particular solution (−533,−2, 110, 261).
1.3 Matrices
Review your notes and the examples there. Then make your own examples up and check them in Matlab.Alternatively, you can Google many examples for the simple operations of matrix addition, subtraction,scalar multiplication and transpose. Also, the many standard texts in Linear Algebra that we have inthe library will also have a multitude of examples covering this basic material.
7) Calculate (A +BT − 2C)T where
A =
−1 21 02 5
, B =
(4 −2 1−1 3 4
), C =
2 −11 23 −3
.Solution:First calculate
A +BT − 2C =
−1 21 02 5
+
(4 −2 1−1 3 4
)T− 2
2 −11 23 −3
=
−1 21 02 5
+
4 −1−2 31 4
−4 −22 46 −6
=
−1 + 4− 4 2− 1 + 21− 2− 2 0 + 3− 42 + 1− 6 5 + 4 + 6
=
−1 3−3 −1−3 15
.Thus
(A +BT − 2C)T =
−1 3−3 −1−3 15
T =
(−1 −3 −33 −1 15
).
†I used the ‘rref’ command in Matlab and typed ‘format rat’ beforehand.
6
1.4 Matrix Multiplication
See the examples in your notes, in standard textbooks, and on the web. This is another type of problemwhere you can readily construct your own examples and easily check them using Matlab. Keep in mindthe comments made regarding the dimensions of the matrices concerned on page 38 of your Workbook.
8) Let
A =
(1 2 30 −1 2
), B =
1 20 −11 4
.Find the matrix product AB.
Solution:
AB =
(1 2 30 −1 2
)︸ ︷︷ ︸
2×3
1 20 −11 4
︸ ︷︷ ︸
3×2
=
(1(1) + 2(0) + 3(1) 1(2) + 2(−1) + 3(4)
0(1) + (−1)(0) + 2(1) 0(2) + (−1)(−1) + 2(4)
)=
(4 122 9
)︸ ︷︷ ︸
2×2
9) Let
U =
123
and V =
−214
.Verify that (a) V TU = UTV and (b) (UV T )T = V UT .
Solution:(a)
V TU =(−2 1 4
)︸ ︷︷ ︸1×3
123
︸ ︷︷ ︸3×1
= −2(1) + 1(2) + 4(3) = 12.
And
UTV =(1 2 3
)︸ ︷︷ ︸1×3
−214
︸ ︷︷ ︸
3×1
= 1(−2) + 2(1) + 3(4) = 12 = V TU X.
(b)
UV T =
123
︸ ︷︷ ︸3×1
(−2 1 4
)︸ ︷︷ ︸1×3
=
1(−2) 1(1) 1(4)2(−2) 2(1) 2(4)3(−2) 3(1) 3(4)
=
−2 1 4−4 2 8−6 3 12
︸ ︷︷ ︸
3×3
.
7
Thus
(UV T )T =
−2 1 4−4 2 8−6 3 12
T =
−2 −4 −61 2 34 8 12
.And
V UT =
−214
︸ ︷︷ ︸
3×1
(1 2 3
)︸ ︷︷ ︸1×3
=
−2(1) −2(2) −2(3)1(1) 1(2) 1(3)4(1) 4(2) 4(3)
=
−2 −4 −61 2 34 8 12
︸ ︷︷ ︸
3×3
= (UV T )T X.
10) Find a number k such that Ax = kx, where
A =
(2 11 2
)and x =
(11
).
Solution:
Ax = kx
=⇒(2 11 2
)(11
)= k
(11
)=⇒
(33
)=
(kk
),
i.e., k = 3.
11) Let
A =
(1 2 34 5 6
), x =
789
.Write the matrix product Ax as a sum of the columns of A multiplied by the entries in x.(Hint: use Theorem 1).
Solution:
Ax =
(1 2 34 5 6
)789
=
(1(7) + 2(8) + 3(9)4(7) + 5(8) + 6(9)
)
= 7
(14
)+ 8
(25
)+ 9
(36
).
1.5 Algebraic Properties of Matrix Operations
Recall that the ‘Algebraic Properties of Matrix Operations’ that we refer to are listed in Theorem 2 ofyour notes.
8
12) Let
A =
(1 23 4
)and B =
(0 11 2
).
(a) Calculate A + B and B + A. Is matrix addition commutative? (i.e., is it generally true thatA +B = B +A ?)
(b) Calculate AB and BA. Is matrix multiplication commutative? (i.e., is it generally true thatAB = BA ?)
Solution:(a) We find that
A +B =
(1 34 6
)= B +A.
Yes, this is generally true.
(b) We find that
AB =
(2 54 11
),
but
BA =
(3 47 10
)6= AB,
thus, no, it is NOT generally true that AB = BA.
Note:See properties (1) and the comment on page 61 of our Workbook.
13) Verify the algebraic property (AB)T = BTAT (Property 11) where
A =
(1 3 22 1 −3
), B =
3 −12 41 2
.Solution:Now
AB =
(1 3 22 1 −3
)3 −12 41 2
=
(1(3) + 3(2) + 2(1) 1(−1) + 3(4) + 2(2)
2(3) + 1(2) + (−3)(1) 2(−1) + 1(4) + (−3)(2)
)=
(11 155 −4
)Thus
(AB)T =
(11 155 −4
)T=
(11 515 −4
).
9
But
BTAT =
3 −12 41 2
T(1 3 22 1 −3
)T
=
(3 2 1−1 4 1
)1 23 12 −3
=
(3(1) + 2(3) + 1(2) 3(2) + 2(1) + 1(−3)−1(1) + 4(3) + 2(2) −1(2) + 4(1) + 2(−3)
)=
(11 515 −4
),
which is the same as before X.
14) Determine the values of the number k such that (kA)T (kA) = 1, where
A =
−21−1
.Solution:We use the algebraic properties of matrices (see your notes):
(kA)T (kA) = (kAT )(kA) (Property 12)
= k2(ATA) (Property 5 & 8)
= k2(−2 1 −1
)−21−1
= k2[(−2)2 + 12 + (−1)2]= k26
= 1.
I.e., 6k2 = 1 =⇒ k2 = 1/6 =⇒ k = ±1/√6.
15) Using the algebraic properties of matrices find an expression for (A−B)T .
Solution:
(A−B)T = (A + (−1)B)T (Definition 12)
= AT + ((−1)B)T (Property 10)
= AT + (−1)BT (Property 12)
= AT −BT (Definition 12).
16) Using the algebraic properties of matrices find an expression for (A−B)2.
10
Solution:
(A−B)2 = (A−B)(A−B) (Definition 16)
= (A + (−1)B)(A + (−1)B) (Definition 12)
= A(A + (−1)B) + (−1)B(A + (−1)B) (Property 4)
= A2 + (−1)AB + (−1)BA + (−1)2B2 (Property 4)
= A2 −AB −BA +B2 (Definition 12).
17) Prove that if x1 and x2 are solutions of the linear system Ax = b, then x1 − x2 is a solution of thecorresponding homogeneous system Ax = 0.
Solution:We are told that
Ax1 = b and Ax2 = b (∗)
So we have
A(x1 − x2) = Ax1 −Ax2 (Property 4)
= b− b (using (∗))
= 0 X.
.
18) Prove for an arbitrary matrix A that(AT )2 = (A2)T .
(Hint: use Theorem 2, part (11).)
Solution:Using the hint with A = B yields
(AA)T = ATAT i.e. (A2)T = (AT )2,
as required.
19) Find examples to prove the following statements, where A, B, and C are 2-by-2 matrices:(a) AB = AC does not imply B = C.(b) AB = O (where O is the ‘zero matrix’) does not imply A = O or B = O.(See page 53 of your Workbook.)
Solution:
(a) Choose A =
(1 11 1
), B =
(2 13 3
), and C =
(3 32 1
), and observe
that AB = AC =
(5 45 4
).
(b) Choose A =
(−1 1−1 1
), B =
(1 11 1
), and observe that AB =
(0 00 0
).
Note:Now you may be wondering how I got the answers I did. Well apart from ’guessing’ one can apply some
11
results from the next section (Special Types of Matrices) to help us‡. In case (a) we know that A mustbe singular (no inverse) because if it was nonsingular, then AB = AC implies A−1(AB) = A−1(AC)implies (A−1A)B = (A−1A)C implies IB = IC or B = C, which contradicts what we are trying toprove. You can easily check in our example above that |A| = 0, i.e. A is singular. A similar argumentapplies to (b). For example, if A is nonsingular then AB = O implies A−1(AB) = A−1O implies(A−1A)B = O implies IB = O i.e. B = O, which contradicts what we are trying to prove. Thus Amust be singular, which is easily checked for our example. A similar argument shows that B must besingular, which it is for our example. The rest is just trial and error.
20) Let A be m× p, B be p× q, and C be q × n. Find an expression for((AB)C
)ij.
Hints:
• Recall that (AB)ij =
p∑s=1
aisbsj (see page 37 of your Workbook).
• Let D = AB and consider the expression for (DC)ij (using a ’dummy’ variable k).
Note:This question isn’t typical of the sort of math problems I set. But it gives a taste of what it is like torigorously prove the algebraic properties of matrices. If you are a math/stats/physics student you shouldattempt it!
Solution:Following the hints we have
(AB)ij =
p∑s=1
aisbsj = (D)ij ≡ dij, (∗)
where D = AB. Now consider
((AB)C)ij = (DC)ij =
q∑k=1
dikckj (By definition)
=
q∑k=1
(p∑s=1
aisbsk
)ckj (Use (∗), but replace j with k)
=
q∑k=1
p∑s=1
aisbskckj.
1.6 Special Types of Matrices
21) Consider the matrix equation ACx = b where
A−1 =
(2 1−1 1
), C−1 =
(2 11 2
), and b =
(23
).
It is easily verified that A−1 and C−1 are nonsingular. Without first finding A and C determine x.
‡Of course this doesn’t help you if you haven’t done that section yet, but I’m assuming you do re-read your notes.
12
Solution:Like any equation involving ‘x’ we need to isolate the unknown. We use the algebraic properties ofmatrices to do so as follows:
ACx = b
=⇒ A−1(ACx) = A−1b (multiply both sides on the left by A−1)
=⇒ (A−1A)︸ ︷︷ ︸I
Cx = A−1b (using the algebraic properties of matrices)
=⇒ Cx = A−1b
=⇒ C−1(Cx) = C−1(A−1b) (multiply both sides on the left by C−1)
=⇒ (C−1C)︸ ︷︷ ︸I
x = C−1A−1b (using the algebraic properties of matrices)
=⇒ x = C−1A−1b
=
(2 11 2
)(2 1−1 1
)(23
)(using the given information)
=
(3 30 3
)(23
)(matrix multiplication)
=
(159
)(matrix multiplication)
Note:From now on when doing problems of this type we don’t always show every step of forming an identitymatrix I = AA−1. We also assume that you know that multiplication of a matrix (on the left or theright) by an identity matrix leaves the matrix unchanged.
22) Let
A−1 =
(1 34 0
)and B−1 =
(−1 13 7
).
Find (AB)−1.
Solution:We could find the inverses of the given matrices (A and B), and then find the inverse of the matrixproduct AB to get (AB)−1. However, there is a much easier way to calculate this. Recall
(AB)−1 = B−1A−1 (Theorem 4 (i))
=
(−1 13 7
)(1 34 0
)=
(3 −331 9
).
23) Consider the linear system ATx = b where
A−1 =
(4 11 0
)and b =
(1−2
).
Find the solution x.
13
Solution:
ATx = b
=⇒ x = (AT )−1b (multiply both sides by (AT )−1 on the left)
= (A−1)T b (Theorem 4(ii))
=
(4 11 0
)T (1−2
)=
(4 11 0
)(1−2
)=
(21
).
24) Prove for an arbitrary square matrix A that
(A2)−1 = (A−1)2.
(Hint: use Theorem 4 (i).)
Solution:Using the hint with A = B yields
(AA)−1 = A−1A−1 i.e. (A2)−1 = (A−1)2,
as required.
25) Use the following invertible matrix
A =
1 −2 2−1 1 31 −1 −4
to encode the message
“ MEET ME MONDAY”
The inverse of the ‘encoding matrix’ above is the following ‘decoding matrix’:
A−1 =
−1 −10 −8−1 −6 −50 −1 −1
Also show how to decode the encoded message.
Solution:As in Application 5, assign a number for each letter of the alphabet and assign 27 to spaces between(and for a space at the end of a word - you will see why in a minute):
M E E T ∗ M E ∗ M O N D A Y ∗13 5 5 20 27 13 5 27 13 15 14 4 1 25 27
As A is 3× 3 break the enumerated message up into sequences of 3× 1 vectors and use these to makethe columns of a new matrix B:
B =
15 20 5 15 15 27 27 14 255 13 13 4 27
14
Now form the matrix product
C = AB =
1 −2 2−1 1 31 −1 −4
15 20 5 15 15 27 27 14 255 13 13 4 27
=
15 −8 −23 −5 55 46 61 11 105−10 −59 −74 −15 −132
.The columns of this matrix give the encoded message in the following linear form:
15, 5, −10, −8, 46, −59, −23, 61, −74−5, 11, −15, 5, 105, −132
To decode the message using the inverse matrix (sent to the receiver of the message separately), wewrite this string as a sequence of 3× 1 vectors (yielding C), and observe that
C = AB =⇒ B = A−1C =
−1 −10 −8−1 −6 −50 −1 −1
15 −8 −23 −5 55 46 61 11 105−10 −59 −74 −15 −132
=
15 20 5 15 15 27 27 14 255 13 13 4 27
X
The numbers written column-wise give the original enumerated message.
26) Consider an arbitrary square nonsingular matrix A and a nonzero scalar c. Prove that
(cA)−1 =1
cA−1
Hint:Recall that
A−1 = B ⇐⇒ AB = I,
where I is the identity matrix. (Note: of course in the hint, A and B are generic matrices, so the Ahere is different from the A in the statement to be proved.)
Solution:Using the hint observe that
(cA)
(1
cA−1
)=
(c ·
1
c
)(AA−1) (Using algebraic matrix properties)
= 1 · I (c/c = 1 and AA−1 = I)
= I X
15
If we want to be pedantic, we could spell it out: in the hint take A to be cA and B to be 1cA−1, etc.
Chapter 2
Solving Linear Systems
2.1 Echelon Form of a Matrix
27) Reduce
A =
6 3 −4−4 1 −61 2 −5
to REF and then to RREF. (Note: number of rows (m) = number of columns (n).)
Solution:Note that the math is simpler if a11 is equal to 1.
A −→
1 2 −5−4 1 −66 3 −4
r1 ↔ r3
−→
1 2 −50 9 −260 −9 26
r2 + 4r1 → r2r3 − 6r1 → r3
−→
1 2 −50 9 −260 0 0
r3 + r2 → r3
−→
1 2 −50 1 −26/90 0 0
r2 · (1/9)→ r2 (REF)
−→
1 0 7/90 1 −26/90 0 0
r1 − 2r2 → r1(RREF).
28) Reduce
A =
1 11 −1−1 2
to REF and then to RREF. (Note: number of rows (m) > number of columns (n).)
16
Solution:
A −→
1 10 −20 3
r2 − r1 → r2r3 + r1 → r3
−→
1 10 10 1
r2 · (−1/2)→ r2r3 · (1/3)→ r3
−→
1 10 10 0
r3 − r2 → r3
(REF)
−→
1 00 10 0
r1 − r2 → r1(RREF).
29) Reduce
A =
(1 2 1 12 4 2 3
)to REF and then to RREF. (Note: number of rows (m) < number of columns (n).)
Solution:
A −→(1 2 1 10 0 0 1
)r2 − 2r1 → r2
(REF)
−→(1 2 1 00 0 0 1
)r1 − r2 → r1 (RREF)
30) Determine the reduced row echelon form of(cosx sinx− sinx cosx
)for 0 < x < π
2.
Show all the steps in your argument!
Solution:Before we start we provide two reminders:
tanx =sinx
cosx, sin2 x + cos2 x = 1.
We use elementary row operations as usual.
Method 1:
(cosx sinx− sinx cosx
)−→
(1 tanx
− sinx cosx
)r1 · 1
cosx→ r1
−→(1 tanx
0 cosx + sin2 xcosx
)r2 + sinx · r1 → r2
17
After observing that
cosx +sin2 x
cosx=
cos2 x + sin2 x
cosx=
1
cosx,
we have (1 tanx
0 cosx + sin2 xcosx
)=
(1 tanx
0 1cosx
)−→
(1 tanx0 1
)r2 · cosx→ r2
−→(1 00 1
)r1 − tanx · r2 → r1
In the past though I have seen students take a more non-standard approach to solving this problem, asthe next method shows.
Method 2:
(cosx sinx− sinx cosx
)−→
(cos2 x cosx sinxsin2 x − sinx cosx
)r1 · cosx→ r1
r2 · (− sinx)→ r2
−→(
1 0sin2 x − sinx cosx
)r1 + r2 → r1
−→(1 00 − sinx cosx
)r2 − sin2 x · r1 → r2
−→(1 00 1
)r2 · 1
(− sinx cosx)→ r2
2.2 Solving Linear Systems
31) Solve the system 2x − 3z = 1
4x + y − 2z = 2
3x + y − z = 3
via (a) Gaussian Elimination, and (b) Gauss-Jordon Elimination. (Note: number of equations (m) =number of unknowns (n).)
Solution:(a) First write down the associated matrix equation:2 0 −3
4 1 −23 1 −1
xyz
=
123
,
18
so the augmented matrix is2 0 −3 14 1 −2 23 1 −1 3
−→
2 0 −3 11 0 −1 −13 1 −1 3
r2 − r3 → r2
−→
1 0 −1 −12 0 −3 13 1 −1 3
r1 ↔ r2
−→
1 0 −1 −10 0 −1 30 1 2 6
r2 − 2r1 → r2r3 − 3r1 → r3
−→
1 0 −1 −10 1 2 60 0 −1 3
r2 ↔ r3
−→
1 0 −1 −10 1 2 60 0 1 −3
r3 · (−1)→ r3
(REF) (∗)
The associated (equivalent) linear system is:x − z = −1y + 2z = 6
z = −3
Back-substitution yields
z = −3, y = 12, x = −4.
(b) Alternatively, from (∗)
(∗) −→
1 0 0 −40 1 0 120 0 1 −3
r1 + r3 → r1r2 − 2r3 → r2 (RREF).
The associated (equivalent) linear system is:x = −4y = 12
z = −3
as before.
32) Solve the system 3x1 + 4x2 = 1
x1 − 2x2 = 2
−x1 + 5x2 = 0
19
via (a) Gaussian Elimination, and (b) Gauss-Jordon Elimination. (Note: number of equations (m) >number of unknowns (n).)
Solution:(a) The augmented matrix is 3 4 1
1 −2 2−1 5 0
leading to
−→
1 −2 23 4 1−1 5 0
r1 ↔ r2
−→
1 −2 20 10 −50 3 2
r2 − 3r1 → r2r3 + r1 → r3
−→
1 −2 20 2 −10 3 2
r2 · (1/5)→ r2
−→
1 −2 20 2 −10 1 3
r3 − r2 → r3
−→
1 −2 20 1 30 2 −1
r2 ↔ r3
−→
1 −2 20 1 30 0 −7
r3 − 2r2 → r3
−→
1 −2 20 1 30 0 1
r3 · (−1/7)→ r3
(REF) (∗)
The associated system of equations is: x1 − 2x2 = 2
x2 = 3
0 = 1
The last equation is impossible, so the system is inconsistent.
(b) We could continue to get (∗) into RREF, i.e.1 0 00 1 00 0 1
(check!), but clearly there is no point.
33) Solve the system
20
{x1 + x2 + 2x3 = 4
2x1 + 3x2 − x3 = 1
via (a) Gaussian Elimination, and (b) Gauss-Jordon Elimination. (Note: number of equations (m) <number of unknowns (n).)
Solution:(a) The augmented matrix is(
1 1 2 42 3 −1 1
)leading to
−→(1 1 2 40 1 −5 −7
)r2 − 2r1 → r2
(REF) (∗)
The associated system of equations is: {x1 + x2 + 2x3 = 4
x2 − 5x3 = −7
The last equation yields x2 = 5x3 − 7, where x3 is free to be chosen. Let x3 = α ∈ R, thenx2 = 5α− 7. From the first equation
x1 = −x2 − 2x3 + 4
= −(5α− 7)− 2α + 4
= −7α + 11
Thus the infinite solution set is
{(−7α + 11, 5α− 7, α) | α ∈ R}
(b) Alternatively, using Gauss-Jordon Elimination we continue the elimination via:
(∗) −→(1 0 7 110 1 −5 −7
)r1 − r2 → r1 (RREF).
The associated linear system is {x1 + 7x3 = 11
x2 − 5x3 = −7
From the 2nd equation x2 = 5x3 − 7. Set x3 = α ∈ R, so x2 = 5α− 7, and from the 1st equationx1 = −7x3 + 11 = −7α + 11, etc. as before.
34) Investigate for what values of a ∈ R the linear system{x + y = 3,
x + (a2 − 8)y = a,
has (i) no solution, (ii) an infinite number of solutions, and (iii) a unique solution.
21
Strategy: This is a harder problem, but just apply Gauss-Jordon Elimination to the associated augmentedmatrix and see what conditions have to be given to the number a for the three cases.
Solution:The eliminations process applied to the augmented matrix initially yields:(
1 1 31 a2 − 8 a
)−→
(1 1 30 a2 − 9 a− 3
)r2 − r1 → r2
(∗)
Now in order to proceed with the GJE algorithm we would divide the 2nd row by a2− 9. However, thiscan only be done if a2 − 9 6= 0, i.e. a 6= ±3. Thus (for now) we exclude these two possibilities of a.Continuing
(∗) −→(1 1 3
0 1 1a+3
)r2 ·
(1
a2−9
)→ r2
−→(1 0 3− 1
a+3
0 1 1a+3
)r1 − r2 → r1,
where in the 1st step above we used the fact that
a− 3
a2 − 9=
����(a− 3)
����(a− 3)(a + 3)=
1
a + 3(as a 6= 3).
Thus we have the unique solution for each a 6= ±3 given by
x = 3−1
a + 3, y =
1
a + 3.
Now we go back and deal with the two cases a = +3 and a = −3:
a = +3:From (∗) we get the augmented matrix (
1 1 30 0 0
)i.e., x + y = 3 so x = 3 − y, where y is free to be chosen, etc. Thus in this case we get an infinitenumber of solutions.
a = −3:From (∗) we get the augmented matrix (
1 1 30 0 −6
).
From the last equation we have 0 = −6, an impossibility, thus the system is inconsistent. I.e., in thiscase we have no solutions.
2.3 Finding A−1
35) Find the inverse of
A =
−3 −3 −40 1 14 3 4
22
(if it exists).
Solution:
[A|I] =
−3 −3 −4 1 0 00 1 1 0 1 04 3 4 0 0 1
−→
1 0 0 1 0 10 1 1 0 1 04 3 4 0 0 1
r1 + r3 → r1
−→
1 0 0 1 0 10 1 1 0 1 00 3 4 −4 0 −3
r3 − 4r1 → r3
−→
1 0 0 1 0 10 1 1 0 1 00 0 1 −4 −3 −3
r3 − 3r2 → r3
−→
1 0 0 1 0 10 1 0 4 4 30 0 1 −4 −3 −3
r2 − r3 → r2
Thus
A−1 =
1 0 14 4 3−4 −3 −3
36) Find the inverse of
A =
1 2 −31 −2 15 −2 −3
(if it exists).
Solution:
[A|I] =
1 2 −3 1 0 01 −2 1 0 1 05 −2 −3 0 0 1
−→
1 2 −3 1 0 00 −4 4 −1 1 00 −12 12 −5 0 1
r2 − r1 → r2r3 − 5r1 → r3
−→
1 2 −3 1 0 00 −4 4 −1 1 00 0 0 −2 −3 1
r3 − 3r2 → r3
.
Thus we see that the original matrix A is not equivalent to the identity matrix (the last row is all zeros)and thus from Theorem 6 the matrix A must be singular (i.e., it does not have an inverse).
37) Investigate for what values of a ∈ R the homogeneous linear system{(a− 1)x + 2y = 0,
2x + (a− 1)y = 0,
23
has a nontrivial solution. Do this problem three different ways, using the contrapositive of items 1., 3.,and 5. with item 2. in Theorem 6.
Solution:
The contrapositive of item 5 with item 2 in Theorem 6 yields:
Ax = 0 has a non-trivial solution ⇐⇒ |A| = 0.
Writing the given system in matrix form(a− 1 2
2 a− 1
)(xy
)=
(00
),
or Ax = 0. So to show that we have a non-trivial solution (i.e., an infinite number of solutions) wemust have that |A| = 0. We solve
|A| =∣∣∣∣a− 1 2
2 a− 1
∣∣∣∣ = (a− 1)2 − 4 = 0
=⇒ (a− 1)2 = 4
=⇒ a− 1 = ±2=⇒ a = 3 or − 1.
The contrapositive of item 3 with item 2 in Theorem 6 yields:
Ax = 0 has a non-trivial solution ⇐⇒ the RREF of A is NOT In.
Remember, to get a non-trivial solution, during the row elimination procedure we would get a row ofzeros, leading to a free variable etc. Writing the given system in matrix form and applying row operations:(
a− 1 22 a− 1
)−→
(2 a− 1
a− 1 2
)r1 ↔ r2
−→(
1 12(a− 1)
a− 1 2
)r1 · (1/2)→ r1
−→(1 1
2(a− 1)
0 2− 12(a− 1)2
)r2 − (a− 1) · r1 → r2
.
We thus have a non-trivial solution provided (for a row of zeros)
2−1
2(a− 1)2 = 0 =⇒ 4− (a− 1)2 = 0
=⇒ (a− 1)2 = 4, etc., as before.
The contrapositive of item 1 with item 2 in Theorem 6 yields:
Ax = 0 has a non-trivial solution ⇐⇒ A is singular.
So we try and find the inverse of A and in the process find conditions telling us that the inverse does
24
not exist:
[A|I] =(a− 1 2 1 0
2 a− 1 0 1
)−→
(2 a− 1 0 1
a− 1 2 1 0
)r1 ↔ r2
−→(
1 12(a− 1) 0 1
2a− 1 2 1 0
)r1 · (1/2)→ r1
−→(1 1
2(a− 1) 0 1
20 2− 1
2(a− 1)2 1 − 1
2
).
So A−1 does not exist if (see Exercise 36))
2−1
2(a− 1)2 = 0 etc. X
Of course we already saw from the previous argument that A cannot be row reduced to I so we didn’treally need to do this calculation.
Chapter 3
Determinants
3.1 Definition
38) Find |A| where
A =
−2 3 21 2 −14 2 18
Solution:We use Definition 29 in the Workbook and use a diagram as explained on page 106 (arrows omitted):
−2 3 2 −2 31 2 −1 1 24 2 18 4 2
Then
|A| = (−2)(2)(18) + (3)(−1)(4) + (2)(1)(2)
− (1)(3)(18)− (−2)(−1)(2)− (2)(2)(4)
= −72− 12 + 4− 54− 4− 16
= −154.
25
3.2 Properties of Determinants
39) Let
A =
−1/2 3 1/31/4 2 −1/61 2 3
Use Theorem 9 (b) and Definition 29 in the Workbook to evaluate |A|.Solution:
|A| =
∣∣∣∣∣∣−1/2 3 1/31/4 2 −1/61 2 3
∣∣∣∣∣∣ = (1/4)
∣∣∣∣∣∣−2 3 1/31 2 −1/64 2 3
∣∣∣∣∣∣ c1 · 4→ c1
= (1/4)(1/6)
∣∣∣∣∣∣−2 3 21 2 −14 2 18
∣∣∣∣∣∣ c3 · 6→ c3
= (1/24)(−154) (Using Exercise 38))
= −77/12.
40) Let
A =
−2 3 21 2 −14 2 18
Use row operations to reduce |A| to upper triangular form (using Theorem 9) and then use Theorem8(c) to evaluate this determinant.
Solution:
|A| = −
∣∣∣∣∣∣1 2 −1−2 3 24 2 18
∣∣∣∣∣∣r1 ↔ r2
= −
∣∣∣∣∣∣1 2 −10 7 00 −6 22
∣∣∣∣∣∣ r2 + 2r1 → r2r3 − 4r1 → r3
= −7
∣∣∣∣∣∣1 2 −10 1 00 −6 22
∣∣∣∣∣∣ r2 · (1/7)→ r2
= −7
∣∣∣∣∣∣1 2 −10 1 00 0 22
∣∣∣∣∣∣ r3 + 6r2 → r3
= (−7)(1)(1)(22) = −154,
as we found in Exercise 38).
41) Let
A =
∣∣∣∣2 −31 2
∣∣∣∣26
(a) Compute |A| and |A−1|.(b) Make a conjecture about the determinant of the inverse of a matrix.
Solution:(a) |A| = 2(2)− 3(−1) = 4 + 3 = 7. Then using the formula for the inverse of a 2× 2 matrix (page69 of the workbook) we have
A−1 =1
|A|
(2 3−1 2
)=
(2/7 3/7−1/7 2/7
).
Thus|A−1| = (2/7)2 − (3/7)(−1/7) = (4 + 3)/49 = 1/7.
(b) We see that |A−1| = 1|A| .
42) Let A and B be square matrices of order 3 such that |A| = 4 and |B| = 5.(a) Find |AB|(b) Find |2A|(c) Are A and B singular or nonsingular? Explain.(d) If A and B are nonsingular find |A−1| and |B−1|(e) Find |(AB)T |Solution:(a) Using Theorem 10
|AB| = |A||B| = 4(5) = 20.
(b)
|2A| =
∣∣∣∣∣∣2(·) 2(·) 2(·)2(·) 2(·) 2(·)2(·) 2(·) 2(·)
∣∣∣∣∣∣= (2)(2)(2)
∣∣∣∣∣∣(·) (·) (·)(·) (·) (·)(·) (·) (·)
∣∣∣∣∣∣ (Using Theorem 9(b) three times)
= 8|A| = 8(4) = 32.
(c) From Theorem 11|A|, |B| 6= 0 ⇐⇒ A and B are nonsingular.
(d) From Corollary 1 (to Theorem 10) we have (after noting the conclusion of (c))
|A−1| =1
|A|=
1
4, |B−1| =
1
|B|=
1
5.
(e)
|(AB)T | = |BTAT | (Using Theorem 2 (part 11))
= |BT ||AT | (Using Theorem 10)
= |B||A| (Using Theorem 7)
= 5(4) = 20.
27
43)
If |A| =
∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3
∣∣∣∣∣∣ = 2,
find |B| =
∣∣∣∣∣∣(3a1 − 6a3) a2 a3(3b1 − 6b3) b2 b3(c1 − 2c3)
13c2
13c3
∣∣∣∣∣∣ .Hint: apply row and column operations (see Theorem 9). This is a harder problem.
Solution:The method here is to apply the rules of Theorem 9 to convert |A| into |B|, but keeping track of howthe operations effect the determinants.
2 =
∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3
∣∣∣∣∣∣ = 1
3
∣∣∣∣∣∣3a1 a2 a33b1 b2 b33c1 c2 c3
∣∣∣∣∣∣ 3c1 → c1
=1
3
∣∣∣∣∣∣(3a1 − 6a3) a2 a3(3b1 − 6b3) b2 b3(3c1 − 6c3) c2 c3
∣∣∣∣∣∣ c1 − 6c3 → c1
= 3(1/3)
∣∣∣∣∣∣(3a1 − 6a3) a2 a3(3b1 − 6b3) b2 b3(c1 − 2c3) (1/3)c2 (1/3)c3
∣∣∣∣∣∣ 13 · r3 → r3.
I.e. we have |B| = 2.
3.3 Cofactor Expansions
44) Let
A =
1 −3 40 2 56 −1 7
Find |M23| (the minor of the entry a23) and A23 (the cofactor of a23).
Solution:Crossing out row 2 and column 3 and taking the determinant of the resulting matrix yields:
|M23| =∣∣∣∣1 −36 −1
∣∣∣∣ = (1)(−1)− (6)(−3) = −1 + 18 = 17.
The cofactor of a23 (signed minor) is then
A23 = (−1)2+3|M23| = −17.
Caution: M23 is a matrix and A23 is a scalar!
45) Let A be the same as in Exercise 44). Use a cofactor expansion along row 2 to evaluate |A|.
28
Solution:
|A| = a21A21 + a22A22 + a23A23
= 0(−1)2+1|M21| + 2(−1)2+2|M22| + 5(−1)2+3|M23|
= 0 + 2
∣∣∣∣1 46 7
∣∣∣∣− 5
∣∣∣∣1 −36 −1
∣∣∣∣= 2(7− 24)− 5(−1 + 18)
= 2(−17)− 5(17)
= −119.
46) Let
A =
(a bc d
).
Use a cofactor expansion (along row 1 or column 2) to prove the usual formula for the determinant ofA, namely
|A| = ad− bc.
Solution:
A =
(a bc d
)pattern of signs:
+ −− +
.
Now do a cofactor expansion along row 1:
|A| = +a|d| − b|c| = ad− bc X
Note:(Here | · | does not represent absolute value, but a determinant of order 1 (see Definition 28).
Alternatively, expanding along column 2:
|A| = −b|c| + d|a| = ad− bc X
47) Let
A =
a b cd e fg h i
.Use a cofactor expansion to verify
|A| = |AT |.
Solution:
A =
a b cd e fg h i
pattern of signs:+ − +− + −+ − +
29
Expanding along the 1st row:
|A| = +a
∣∣∣∣e fh i
∣∣∣∣− b ∣∣∣∣d fg i
∣∣∣∣ + c ∣∣∣∣d eg h
∣∣∣∣But
AT =
a d gb e hc f i
pattern of signs:+ − +− + −+ − +
Expanding along the 1st column:
|AT | = +a
∣∣∣∣e hf i
∣∣∣∣− b ∣∣∣∣d gf i
∣∣∣∣ + c ∣∣∣∣d ge h
∣∣∣∣= +a
∣∣∣∣e fh i
∣∣∣∣− b ∣∣∣∣d fg i
∣∣∣∣ + c ∣∣∣∣d eg h
∣∣∣∣= |A|,
where we used the fact that for an arbitrary determinant of order 2:∣∣∣∣a bc d
∣∣∣∣ = ad− bc = ad− cb =
∣∣∣∣a cb d
∣∣∣∣ .
3.4 Inverse of a Matrix (via the Adjoint)
48) Find the adjoint of
A =
−1 3 20 −2 11 0 −2
.
Solution:Reminder: the cofactor of a12 is
A12 = (−1)1+2
∣∣∣∣0 11 −2
∣∣∣∣ = −(0− 1) = 1.
Finding the matrix of cofactors yields (exercise):A11 A12 A13
A21 A22 A23
A31 A32 A33
=
4 1 26 0 37 1 2
.The transpose of this matrix is the adjoint of A, that is:
adj(A) =
4 6 71 0 12 3 2
.
30
49) Using adj(A), where A is the matrix in Exercise 48), find A−1.
Solution:
|A| =
∣∣∣∣∣∣−1 3 20 −2 10 3 0
∣∣∣∣∣∣ r3 + r1 → r3
= −3∣∣∣∣−1 20 1
∣∣∣∣ (Cofactor expansion along row 3)
= −3(−1− 0)
= 3.
Thus recalling the matrix adj(A) from Exercise 48) we have that
A−1 =1
|A|adj(A)
=1
3
4 6 71 0 12 3 2
=
4/3 2 7/31/3 0 1/32/3 1 2/3
.50) Let
A =
(a bc d
)be a nonsingular matrix. Use the formula
A−1 =1
|A|adj(A)
to verify the formula
A−1 =1
ad− bc
(d −b−c a
)Solution:We know that |A| = ad− bc, so all we need to show is that
adj(A) =
(d −b−c a
).
From our notes we know that
adj(A) =
(A11 A12
A21 A22
)T.
Now with
A =
(a bc d
)pattern of signs:
+ −− +
31
we have
A11 = +|d| = +d,
A12 = −|c| = −c,A21 = −|b| = −b,A22 = +|a| = +a,
thus
adj(A) =
(A11 A12
A21 A22
)T=
(d −c−b a
)T=
(d −b−c a
)X
(Here | · | does not represent absolute value, but a determinant of order 1 (see Definition 28).
Chapter 4
Real Vector Spaces
4.1 Vectors in the Plane and in 3-Space (generalized to n-Space)
51) Consider the vectors u = (3,−4), v = (9, 1), and w = (−39, 0).
(a) Use directed line segments to represent u and v.
(b) Find u + v and represent graphically.
(c) Find 2v − u and represent graphically.
(d) Write the vector w as a linear combination of u and v.
Solution:(a) See the figure below.(b) u + v = (3,−4) + (9, 1) = (12,−3). See the figure below.(c) 2v − u = 2(9, 1)− (3,−4) = (18, 2)− (3,−4) = (15, 6). See the figure below.(d) We seek x and y s.t.
xu + yv = w
i.e. x(3,−4) + y(9, 1) = (−39, 0),
or {3x + 9y = −39−4x + y = 0
or
32
{x + 3y = −13
−4x + y = 0
which yields (exercise) x = −1 and y = −4.
52) (a) Draw the vectors
~OP =
(42
), ~OR =
(33
), ~OQ =
(26
).
(b) How do we get to R using the vectors ~OP and ~OQ?
Solution:(a) See the figure below:
33
(b) We seek x and y such that
~OR = x · ~OP + y · ~OQ (∗)
i.e. (33
)= x
(42
)+ y
(26
),
or {4x + 2y = 3
2x + 6y = 3
Solving these equations yields (exercise) x = 3/5, y = 3/10, so from (∗):
~OR =3
5· ~OP +
3
10· ~OQ,
illustrated in red above.
53) Let x = (−1,−2,−2), u = (0, 1, 4), v = (−1, 1, 2) and w = (3, 1, 2) be vectors in R3. Write (ifpossible) x as a linear combination of the vectors u, v and w. (In other words, write x as a sum ofconstants times u, v and w.)
Solution:We seek constants a, b, c ∈ R such that
x = au + bv + cw (∗)
(There may or may not be constants a, b and c such that (∗) holds.) I.e., we want
(−1,−2,−2) = a(0, 1, 4) + b(−1, 1, 2) + c(3, 1, 2)= (0, a, 4a) + (−b, b, 2b) + (3c, c, 2c) (Using Definition 36)
= (−b + 3c, a + b + c, 4a + 2b + 2c) (Using Definition 36 again)
Equating corresponding components on both sides:
−b + 3c = −1 (1st components)
a + b + c = −2 (2nd components)
4a + 2b + 2c = −2 (3rd components)
We have 3 linear equations in 3 unknowns. Applying elementary row operations to get the associatedaugmented matrix into upper triangular form yields0 −1 3 −1
1 1 1 −24 2 2 −2
−→ · · · −→
1 1 1 −20 −1 3 −10 0 −4 4
.(Exercise: Apply the operations: r1 ↔ r2, r3 − 4r1 → r1, r3 · (1/2)→ r3, and r3 − r2 → r3.)
The associated linear system is
34
a + b + c = −2− b + 3c = −1− 4c = 4
Thus applying back-substitution yields
c = −1 =⇒ b = −2 =⇒ a = 1.
Thus from (∗)x = u− 2v − w.
4.2 Vector Spaces
54) Let
V = the set of integers with the standard operations
of (vector) addition and (scalar) multiplication.
Show that V is NOT a vector space.
Solution:All we need is a single counter-example. We show that V is not closed with respect to scalar multipli-cation. I.e., if we multiply a member of V by a scalar then we don’t necessarily get an integer as ananswer (see the comments on page 150 of our notes). Now we are dealing with real vector spaces whichmeans the scalars can be any number in R. So just observe, e.g.:
1
2︸︷︷︸scalar
· (1)︸︷︷︸integer
=1
2︸︷︷︸non-integer
According to Definition 37, V is clearly not closed with respect to scalar multiplication (as defined inthis system) and so V is not a vector space.
Note: however, V is closed with respect to (vector) addition as
integer + integer = integer,
always!
55)
V = the set of all second-degree polynomials.
Show that V is NOT a vector space.
Solution:Once again we find a single counter-example. We show that V is not closed with respect to (vector)addition. In other words, we find two polynomials such that when they are added together (using theusual method of adding polynomials together) we obtain a polynomial that is not of second degree.
Considerp(x) = x2, and q(x) = −x2 + x + 1.
35
Then observep(x) + q(x) = x + 1,
which is a first-degree polynomial. Thus V is not closed with respect to (vector) addition and soaccording to Definition 37 V is not a vector space.
56) LetV = {(x, x− 2) ∈ R2 | x ∈ R}.
Prove that V is NOT a vector space.
Solution:There are various ways we can answer this question. Let’s start with the closure properties that a vectorspace must have. (Note: without any other information we must assume that the operations of (vector)addition and scalar multiplication are inherited from the usual ones for R2.)
Vector additionConsider two arbitrary members of V , namely
v1 = (x1, x1 − 2) ∈ V, v2 = (x2, x2 − 2) ∈ V, x1, x2 ∈ R.
Adding them together:
v1 + v2 = (x1, x1 − 2) + (x2, x2 − 2)
= (x1 + x2, x1 + x2 − 4)
= (x, x− 4)︸ ︷︷ ︸/∈V
6= (x, x− 2),
where x := x1+x2. In other words when we add two arbitrary members of V together we get somethingwhich doesn’t belong to V , i.e. V is not closed with respect to addition. And hence V is not a vectorspace according to Definition 37.
We could stop here. But just for clarity let’s look at two other ways we could arrive at the sameconclusion.
Scalar multiplicationConsider an arbitrary member of V and an arbitrary scalar:
v = (x, x− 2) ∈ V, k ∈ R.
Now do scalar multiplication:
kv = k(x, x− 2)
= (kx, kx− 2k)
= (x, x− 2k)︸ ︷︷ ︸/∈V
6= (x, x− 2) unless k = 1,
(x := kx). In other words scalar multiplication of a member of V doesn’t produce something belongingto V (unless the scalar is equal to 1). And hence V is not a vector space.
Existence of a zero vectorAccording to Definition 37 (item A.2) if V is a vector space it must possess a zero vector. Suppose forthe moment that V does have a zero vector, denoted 0. Then there must exist x0 ∈ R such that
0 = (x0, x0 − 2) ∈ V,
36
so that for any member (x, x− 2) ∈ V we have
(x, x− 2) + (x0, x0 − 2) = (x, x− 2) (See item A.2),
i.e.(x + x0, x + x0 − 4) = (x, x− 2).
Equating corresponding coefficients on both sides:
x + x0 = x (1st components)
x + x0 − 4 = x− 2 (2nd components)
From the 1st equation we getx0 = 0.
But from the 2nd equation we get
x0 = (x− 2)− x + 4 = 2,
and so the equations are inconsistent, and thus there is no x0 that will yield a zero vector. Hence V isnot a vector space.
57) Verify that the standard examples in our notes (Examples 10 - 13) are vector spaces by checking thatall the axioms of Definition 37 hold.
Solution:This is a tedious exercise that is omitted from these solutions. However, you should still work throughthem on your own! If you get stuck get help at the Math & Stats Learning Centre.
4.3 Subspaces
58) Show thatW = {(x, 0, z) ∈ R3 | x, z ∈ R}
is a subspace of R3 with the standard operations.
Solution:Note: W can be interpreted as the xz-plane:
Closure with respect to addition ?:Consider
w1 = (x1, 0, z1) ∈W, w2 = (x2, 0, z2) ∈W,
then
w1 + w2 = (x1, 0, z1) + (x2, 0, z2)
= (x1 + x2, 0, z1 + z2)
= (x3, 0, z3) ∈W X,
where x3 := x1 + x2 and z3 := z1 + z2. I.e., W is closed under (vector) addition.
Closure with respect to scalar multiplication ?:Consider
w = (x, 0, z) ∈W, k ∈ R,
37
then
kw = k(x, 0, z)
= (kx, 0, kz)
= (x, 0, z) ∈W X,
where x := kx and z := kz. I.e., W is closed under scalar multiplication. Thus according to Theorem15 W is a subspace of R3 (and hence a vector space in its own right with the same operations as forR3).
59) Which of these two sets is a subspace of R2?a) The set of points on the line x + 2y = 0.b) The set of points on the line x + 2y = 1.
Solution:(a) We are restricted to the vectors on the line y = − 1
2x. I.e., the set
W =
{(x,−
1
2x
)∈ R2 | x ∈ R
}.
Closure with respect to addition ?:Consider
w1 = (x1,−1
2x1) ∈W, w2 = (x2,−
1
2x2) ∈W,
then
w1 + w2 = (x1,−1
2x1) + (x2,−
1
2x2)
= (x1 + x2,−1
2(x1 + x2))
= (x3,−1
2x3) ∈W X,
where x3 := x1 + x2, I.e., W is closed under addition.
38
Closure with respect to scalar multiplication ?:Consider
w = (x,−1
2x) ∈W, k ∈ R,
then
kw = k(x,−1
2x)
= (kx, k(−1
2x))
= (kx,−1
2(kx))
= (x,−1
2x) ∈W X,
where x := kx, i.e., W is closed under scalar multiplication.
Thus W is a subspace of R2.
(b) We are restricted to vectors on the line y = 12− 1
2x. I.e., the set
W =
{(x,
1
2−
1
2x
)∈ R2 | x ∈ R
}.
Closure with respect to addition ?:Consider
w1 = (x1,1
2−
1
2x1) ∈W, w2 = (x2,
1
2−
1
2x2) ∈W,
then
w1 + w2 = (x1,1
2−
1
2x1) + (x2,
1
2−
1
2x2)
= (x1 + x2, 1−1
2(x1 + x2))
= (x3, 1−1
2x3) /∈W,
where x3 := x1 + x2, I.e., W is NOT closed under addition. We could stop here as all we need is oneof the closure properties to not hold to know that W is not a vector space. For completeness we alsocheck scalar multiplication.
Closure with respect to scalar multiplication ?:Consider
w = (x,1
2−
1
2x) ∈W, k ∈ R,
then
kw = k(x,1
2−
1
2x)
= (kx,k
2−k
2x))
= (kx,k
2−
1
2(kx))
= (x,k
2−
1
2x) /∈W (unless k = 1),
39
where x := kx, i.e., W is NOT closed under scalar multiplication, thus again we have proved that Wis not a subspace of R2. (E.g., taking k = 0 yields kw = (0, 0) which doesn’t lie on the line - seebelow.)
Existence of a zero vector ?Alternatively, we know from lecture notes that if W is a subspace of R2 then W must possess the
same zero vector as R2 (i.e., (0, 0) ≡(00
)), which it clearly doesn’t as the set of points defining W
lie on a straight line that doesn’t pass through the origin. Hence once again we see that W cannot bea subspace of W .
60) Draw diagrams to illustrate the results we found for (vector) addition in Exercise 59). For part (a)choose w1 = (1,− 1
2), w2 = (3,− 3
2), and for part (b) choose w1 = (1, 0) and w2 = (−1, 1).
40
(a)
(b)
Solution:(a) First observe that
w1 + w2 = (1,−1
2) + (3,−
3
2) = (4,−2) ∈W X.
Showing this vector addition on a diagram (see below) we see that the resultant vector w3 remainson the straight line, which illustrates that vector addition is closed. (The parallelogram law for vectoraddition still holds, but with a parallelogram of zero area.)
(b) Observe thatw1 + w2 = (1, 0) + (−1, 1) = (0, 1) /∈W.
Showing this vector addition on a diagram (see below) we see that the resultant vector w3 is not on thestraight line, which illustrates that vector addition is not closed. (The usual parallelogram law is clearlysee.)
41
61) Prove thatW = {(x1, x2) ∈ R2 |x1 ≥ 0 and x2 ≥ 0}
with the standard operations is not a subspace of R2.
Solution:Clearly W is closed under vector addition (two non-negative numbers added together always yields anon-negative number). However, for scalar multiplication observe (with a scalar k = −1):
−1 · (1, 1) = (−1,−1) /∈W,
so W is not closed under scalar multiplication and hence is not a subspace of R2.
Note:remember, to show that a set is not a closed only a single counter example is needed.
62) Prove that the set
2. W =
{(x3x
)∈ R2 |x ∈ R
}with the standard operations is a subspace of R2.
Solution:(i) Let
w1 =
(x1
3x1
)∈W, w2 =
(x2
3x2
)∈W,
so
w1 + w2 =
(x1
3x1
)+
(x2
3x2
)=
(x1 + x2
3(x1 + x2)
)=
(x3
3x3
)∈W,
where x3 := x1 + x2. So W is closed under vector addition.
(ii) Let
w =
(x3x
)∈W, k ∈ R,
then
k
(x3x
)=
(kx
3(kx)
)=
(x3x
)∈W,
where x := kx. So W is closed with scalar multiplication. Hence W is a subspace of R2.
63) Is the set of all vectors of the form ab
a + 2b
where a, b ∈ R, a subspace of R3?
Solution:Closure with respect to addition ?:Just add to vectors of the given form together: a1
b1a1 + 2b1
+
a2b2
a2 + 2b2
=
a1 + a2b1 + b2
a1 + a2 + 2(b1 + b2)
=
a3b3
a3 + 2b3
,42
where a3 := a1 + a2 and b3 := b1 + b2. So yes, the set is closed with respect to addition.
Closure with respect to scalar multiplication ?:Consider k ∈ R times a representative member of the given set:
k
ab
a + 2b
=
kakb
k(a + 2b)
=
kakb
ka + 2(kb)
=
a
b
a + 2b
,where a := ka and b := kb. So yes, the set is closed with respect to scalar multiplication. Thusaccording to Theorem 15 the given set of vectors is a subspace of R3.
64) Let W be the set of all nonsingular 2-by-2 matrices. Prove that W is NOT a subspace of M22 (thevector space of all 2-by-2 matrices with the usual operations)?(Hint: all you need is a single counter-example for the closure property of vector addition, i.e., choosetwo nonsingular matrices whose sum is singular. Recall that we can tell easily if a matrix is singular ornot from Theorem 11.)
Solution:First recall from Theorem 11 that
A nonsingular ⇐⇒ |A| 6= 0.
We prove that W is not closed under (vector) addition. So using the hint we choose two nonsingularmatrices whose sum is singular. Observe that with
A =
(1 00 1
), B =
(−1 00 −1
),
we have |A| = |B| = 1 (i.e., A and B are nonsingular), but
|A +B| =∣∣∣∣0 00 0
∣∣∣∣ = 0.
I.e., A + B is singular. Thus W not closed under addition and so W is not a subspace of M22.
Exercise: replace B with
(0 11 0
)to get the same conclusion.
65) Prove the corresponding result to Exercise 64) for singular matrices.
Solution:First recall from Theorem 11 that
A singular ⇐⇒ |A| = 0,
(the contrapositive of the statement given in Exercise 64). We prove that W is not closed under (vector)addition. So using the hint we choose two singular matrices whose sum is nonsingular. Observe thatwith
A =
(1 00 0
), B =
(0 00 1
),
we have |A| = |B| = 0 (i.e., A and B are singular), but
|A +B| =∣∣∣∣1 00 1
∣∣∣∣ = 1.
I.e., A +B is nonsingular. Thus W not closed under addition and so W is not a subspace of M22.
43
66) Give a geometrical interpretation of why the set of vectors on the unit circle (centred at the origin) isnot a subspace of R2.
Solution:Consider the two vectors on the unit circle:(
10
)and
(01
),
Illustrated below. Adding them together(10
)+
(01
)=
(11
),
yields a vector that is clearly not on the circle. Hence the given subset of vector is not closed undervector addition and so cannot be a subspace of R2. We could also note that the set of vectors on theunit circle do not possess the zero vector for R2 (namely, the origin), and so (by Corollary 4 to Theorem15) it cannot be a subspace.
4.4 Span
67) Consider the following set of vector in R3:
S = {(1, 3, 1)︸ ︷︷ ︸v1
, (0, 1, 2)︸ ︷︷ ︸v2
, (1, 0,−5)︸ ︷︷ ︸v3
}.
Verify that the vector v1 can be written as a linear combination of v2 and v3, in the form v1 = 3v2+v3.
Solution:Just observe
3v2 + v3 = 3(0, 1, 2) + (1, 0,−5)= (1, 3, 1)
≡ v1 X
68) Consider the following set of vectors in M22:
S ={(0 8
2 1
)︸ ︷︷ ︸
v1
,
(0 21 0
)︸ ︷︷ ︸
v2
,
(−1 31 2
)︸ ︷︷ ︸
v3
,
(−2 01 3
)︸ ︷︷ ︸
v4
}.
44
Verify that the vector v1 can be written as a linear combination of v2, v3, and v4, in the form v1 =v2 + 2v3 − v4.
Solution:
v2 + 2v3 − v4 =
(0 21 0
)+ 2
(−1 31 2
)−(−2 01 3
)=
(0 82 1
)≡ v1 X
69) If possible, write the vector w = (1, 1, 1) as a linear combination of the vectors in the set S:
S = {(1, 2, 3)︸ ︷︷ ︸v1
, (0, 1, 2)︸ ︷︷ ︸v2
, (−1, 0, 1)︸ ︷︷ ︸v3
}.
If there is more than one solution find one particular solution.
Solution:We seek scalars c1, c2, and c3 such that
w = c1v1+c2v2+c3v3 (∗)
I.e., we need
(1, 1, 1) = c1(1, 2, 3) + c2(0, 1, 2) + c3(−1, 0, 1)= (c1, 2c1, 3c1) + (0, c2, 2c2) + (−c3, 0, c3)= (c1 − c3, 2c1 + c2, 3c1 + 2c2 + c3).
Equating corresponding components on both sides yields the linear equations:c1 − c3 = 1
2c1 + c2 = 1
3c1 + 2c2 + c3 = 1
Using elementary row operations, the associated augmented matrix of this system row reduces to (Ex-ercise): 1 0 −1 1
0 1 2 −10 0 0 0
So the system has an infinite number of solutions. The associated linear system is{
c1 − c3 = 1
c2 + 2c3 = −1
Starting from the lasr equation we have c2 = −2c3 − 1, where c3 is a free variable. Set c3 = α, soc2 = −2α− 1, and from the first equation we get c1 = α + 1. I.e., we have the infinite solution set
{(α + 1︸ ︷︷ ︸c1
,−2α− 1︸ ︷︷ ︸c2
, α︸︷︷︸c3
) | α ∈ R}.
45
E.g., choosing α = 1 yields the particular solution
(c1, c2, c3) = (2,−3, 1).
So from (∗) we havew = 2v1 − 3v2 + v3.
70) Try and write the vectorw := (1,−2, 2)
as a linear combination of the vectors in the set
S = {(1, 2, 3)︸ ︷︷ ︸v1
, (0, 1, 2)︸ ︷︷ ︸v2
, (−1, 0, 1)︸ ︷︷ ︸v3
}.
Why can this not be done? (see Exercise 69)).
Solution:Following the procedure from Exercise 69) yields
c1 − c3 = 1
2c1 + c2 = −23c1 + 2c2 + c3 = 2
Using elementary row operations, the associated augmented matrix of this system row reduces to (Ex-ercise): 1 0 −1 0
0 1 2 00 0 0 1
.Clearly from the last row we see that the associated system of equations is inconsistent, thus there isno solution. Consequently W cannot be written as a linear combination of v1, v2 and v3.
71) Consider the following vectors in R3: 123
︸ ︷︷ ︸v1
,
−105
︸ ︷︷ ︸
v2
,
672
︸ ︷︷ ︸v3
.
Using Theorem 1 find an expression for the span (a noun !) of these vectors involving a matrix-vectorproduct.
Solution:
span{v1, v2, v3} = {αv1 + βv2 + γv3 | α, β, γ ∈ R}
=
α123
+ β
−105
+ γ
672
| α, β, γ ∈ R
=
1 −1 62 0 73 5 2
αβγ
| α, β, γ ∈ R
.
46
72) By considering the span (a noun) of the following vectors in R3100
︸ ︷︷ ︸v1
,
010
︸ ︷︷ ︸v2
,
001
︸ ︷︷ ︸v3
,
show that these vectors span (a verb!) R3. (If you haven’t guessed already, this question is more aboutterminology and notation than actual math - the math result is actually quite trivial.)
Solution:We are being asked to show that
span{v1, v2, v3} = R3.
Now
span{v1, v2, v3} ={α
100
+ β
010
+ γ
001
| α, β, γ ∈ R}
={αβ
γ
| α, β, γ ∈ R}.
But α, β, γ are completely arbitrary, so every vector in R3 can be represented in the span of the threevectors.
73) Find sets of vectors that span(a) R2
(b) P2
(c) M22
Solution:(a) The set S = {(1, 0), (0, 1)} spans R2 because any vector (a, b) ∈ R2 can be written as
(a, b) = a(1, 0) + b(0, 1).
(b) The set S = {1, x, x2} spans P2 because any polynomial p(x) = a + bx + cx2 ∈ P2 can bewritten as
p(x) = a(1) + b(x) + c(x2) = a + bx + cx2.
(c) The set
S =
{(1 00 0
),
(0 10 0
),
(0 01 0
),
(0 00 1
)}spans M22 because any
(a bc d
)∈M22 can be written as
(a bc d
)= a
(1 00 0
)+ b
(0 10 0
)+ c
(0 01 0
)+ d
(0 00 1
).
74) Do the vectors 123
,012
,−20
1
47
span R3?
Solution:
Let u =
abc
∈ R3 (for arbitrary a, b, c ∈ R). We seek scalars x, y and z such that
abc
= x
123
+ y
012
+ z
−201
=
1 0 −22 1 03 2 1
xyz
(∗)
(Using Theorem 1). Now ∣∣∣∣∣∣1 0 −22 1 03 2 1
∣∣∣∣∣∣ = −1 6= 0 (exercise)
so the system (∗) has a unique solution for every right hand side vector (a, b, c)T . And as the num-bers a, b and c are arbitrary, we can write every vector in R3 as a linear combination of the given vectors.
Important note:Students often ask me
“what about the case of an infinite number of solutions?”
Remember, if the determinant of the coefficient matrix was equal to zero there are two possibilities -either no solution, or an infinite number of solutions. So it’s natural to wonder for these sorts of problemsif it is sufficient (for the vectors to span the space) just to check if the determinant of the coefficientmatrix is non-zero. The answer is “yes”. Why? Suppose we are going to get an infinite number ofsolutions associated with the augmented matrix of the form∗ ∗ ∗ a
∗ ∗ ∗ b∗ ∗ ∗ c
.For there to be an infinite number of solutions we must have free variables because one or more rowsduring the row reduction procedure are zero. However, this will always lead to a restriction on theparameters a, b and c. E.g., suppose (for the sake of argument) that we apply the row operationsr2 − 3r1 → r2, r3 + 2r1 → r3, and r3 − r2 → (typical sort of operations used to get the matrixin upper triangular form) leading to the last row of the augmented matrix having all zeros. Then theequivalent augmented matrix will be in the form∗ ∗ ∗ a
∗ ∗ ∗ b− 3a0 0 0 c− b + 5a
.And for a consistent set of equations we must impose the restriction c − b + 5a = 0. So all vectors(a, b, c)T that don’t lie on this plane (i.e. c − b + 5a 6= 0) will yield an inconsistent system, and socan’t be written as a linear combination of the given vectors.
Well, you asked!
48
4.5 Linear Independence
75) Determine whether the following set of vectors in R3 is linearly independent or linearly dependent:
S =
{123
︸ ︷︷ ︸v1
,
012
︸ ︷︷ ︸v2
,
−201
︸ ︷︷ ︸
v2
}.
Solution:We seek scalars α1, α2 and α3 ∈ R such that
α1v1+α2v2+α3v3 = 0, (∗)
i.e.,
α1
123
+ α2
012
+ α3
−201
=
000
i.e.
1 0 −22 1 03 2 1
α1
α2
α3
=
000
.Applying Gauss-Jordon elimination to the associated augmented matrix yields (exercise):1 0 −2 0
2 1 0 03 2 1 0
−→1 0 0 00 1 0 00 0 1 0
.Thus we have the unique (‘trivial’) solution α1 = α2 = α3 = 0 to (∗) and so the set S is linearlyindependent.
76) Determine whether the following set of vectors in R2 is linearly independent or linearly dependent:
S =
{(24
)︸︷︷︸v1
,
(−3−6
)︸ ︷︷ ︸v2
.
Solution:You may have spotted that these vectors are clearly scalar multiples of each other, e.g. v1 = (−2/3)v2(hence the vectors are l.d.), but the point here is to learn a consistent methodology to find all possiblesolutions, even for more complicated problems involving multiple vectors.
We seek scalars α1 and α3 ∈ R such that
α1v1+α2v2 = 0, (∗)
i.e.,
α1
(24
)+ α2
(−3−6
)=
(00
)i.e.
(2 −34 −6
)(α1
α2
)=
(00
).
49
Applying Gauss-Jordon elimination to the associated augmented matrix yields (exercise):(2 −3 04 −6 0
)−→
(1 −3/2 00 0 0
).
I.e., we have the single equation α1 − (3/2)α2, where α2 is free. Thus we have an infinite number ofsolutions to (∗). E.g. if α2 = 2/3 then α1 = 1, so the particular solution to (∗) is v1 + (2/3)v2 = 0,or v1 = (−2/3)v2, as we (may have) guessed from the start.
77) Determine whether the set of vectors given below is linearly independent or linearly dependent:
S = {1 + x− 2x2︸ ︷︷ ︸v1
, 2 + 5x− x2︸ ︷︷ ︸v2
, x + x2︸ ︷︷ ︸v3
} ⊂ P2.
Solution:The method for polynomials is the same as for other vectors. As usual, we seek c1, c2, c3 ∈ R suchthat∗
c1v1 + c2v2 + c3v3 = 0 (?)
(Here of course we understand that 0 = 0 + 0x + 0x2). So we have
c1(1 + x− 2x2) + c2(2 + 5x− x2) + c3(x + x2) = 0,
or, after a bit of tidying up
(c1 + 2c2) + (c1 + 5c2 + c3)x + (−2c1 − c2 + c3)x2 = 0.
Now equating coefficients of like powers of x on both sides yields:
x0 : c1 + 2c2 = 0x1 : c1 + 5c2 + c3 = 0x2 : −2c1 − c2 + c3 = 0
Writing these equations as a single matrix equation: 1 2 01 5 1−2 −1 1
c1c2c3
=
000
.Applying Gaussian elimination to to the augmented matrix yields (exercise): 1 2 0 0
1 5 1 0−2 −1 1 0
−→1 2 0 00 1 1/3 00 0 0 0
.The last row of zeros tells us we have free variables (more unknowns than equations) so there will bean infinite number of solutions. So the equation (?) must have a non-trivial solution, i.e., the originalvectors are linearly dependent.
78) Determine whether the set of vectors given below is linearly independent or linearly dependent:
S ={(2 1
0 1
)︸ ︷︷ ︸
v1
,
(3 02 1
)︸ ︷︷ ︸
v2
,
(1 02 0
)︸ ︷︷ ︸
v3
}⊂M22
∗I’m getting lazy, so I’m using c’s instead of α’s.
50
Solution:The same method applies. We seek scalars c1, c2, c3 ∈ R such that
c1v1 + c2v2 + c3v3 = 0 (?)
I.e.,
c1
(2 10 1
)+ c2
(3 02 1
)+ c3
(1 02 0
)=
(0 00 0
),
or (2c1 c10 c1
)+
(3c2 02c2 c2
)+
(c3 02c3 0
)=
(0 00 0
),
or (2c1 + 3c2 + c3 c1
2c2 + 2c3 c1 + c2
)=
(0 00 0
).
Equating corresponding entries of the matrices on both sides:2c1 + 3c2 + c3 = 0
c1 = 0
2c2 + 2c3 = 0
c1 + c2 = 0
Note that there is no need to apply row operations to the associated augmented matrix as we can solvethis system easily as it stands. The 2nd equation yields c1 = 0 =⇒ c2 = 0 (using 4th equation)=⇒ c3 = 0 (using the 3rd equation). Thus we see from (?) that the set S is linearly independent.
4.6 Basis and Dimension
Exercise 62 in the Workbook is a key example showing how to verify that a given set of vectors is a basisfor a vector space. Once we know the dimension of the vector space concerned, according to Theorem19, if the number of vectors is equal to the dimension then all we need to check if we have a basis isthat either the set of vectors is linear independence or, the vectors span the space (not both). If thenumber of vectors is not equal to the dimension of the space then the vectors can’t be a basis!
We initially give two simple exercises:
79) Prove that {(34
),
(−10
)}is a basis for R2.
Solution:According to Theorem 19, as dimR2 = 2, and the number of vectors we consider is also 2, we needonly show that the vectors are linearly independent (or that they span the space). Consider linear inde-pendence:
Seek α1 and α2 such that
α1
(34
)+ α2
(−10
)=
(00
),
51
i.e. (3 −14 0
)(α1
α2
)=
(00
).
Now we can show using Gauss-Jordon elimination that(3 −14 0
)−→
(1 00 1
),
(exercise), or ∣∣∣∣3 −14 0
∣∣∣∣ = 4 6= 0,
and so we have the unique (‘trivial solution’) α1 = α2 = 0, i.e., the vectors are linearly independent, andhence by Theorem 19 (i) the vectors are a basis for R2. So we can stop here, but for completeness we dothe argument that also shows that the vectors span R2 (whether you go for showing linear independence,or span is up to you).
We seek α1, α2 such that for any
(ab
)∈ R2 we have
α1
(34
)+ α2
(−10
)=
(ab
),
i.e. (3 −14 0
)(α1
α2
)=
(ab
).
I.e., instead of the homogeneous system Ax = 0 that we got while considering linear independence,we have the associated nonhomogeneous system Ax = b, which has a unique solution (for every righthand side b) as we already showed as A is nonsingular (because |A| 6= 0).
Note:If the given two vectors were not a basis for R2 then we would have found that |A| = 0, indicating thatA is a singular matrix, which tells us that the given two vectors are not linear independent and don’tspan the space. We now look at two examples of bases for vector spaces other than Rn.
80) Show that the setS = {2, x− 1, x2 + 1}
is a basis for P2.
Solution:This may look like a different problem, but it is done in exactly the same way. The dimension of P2
is three, and as we are considering three vectors as a possible basis, according to Theorem 19 to showthat we have a basis, all we need do is show the given vectors are linearly independent (or, span P2).
To check linear independence, we seek c1, c2, c3 ∈ R such that
2c1+c2(x−1)+c3(x2+1) = 0 ≡ 0+0x+0x2. (∗)
This simplifies to:(2c1 − c2 + c3) + c2x + c3x
2 = 0.
Equating corresponding coefficients of like powers on both sides yields:
52
2c1 − c2 + c3 = 0 (x0)
c2 = 0 (x1)
c3 = 0 (x2)
This readily yields c3 = c2 = 0 =⇒ c1 = 0, hence from (∗) we see that the vectors are linearlyindependent. Alternatively, we could have verified that the vectors span P2, as shown below:
We seek c1, c2, c3 ∈ R such that
2c1+c2(x−1)+c3(x2+1) = a0+a1x+a2x
2, (∗)
for fixed, but arbitrary a0, a1, a2 ∈ R. This simplifies to:
(2c1 − c2 + c3 − a0) + (c2 − a1)x + (c3 − a2)x2 = 0.
Equating corresponding coefficients of like powers on both sides yields:2c1 − c2 + c3 = a0 (x0)
c2 = a1 (x1)
c3 = a3 (x2)
Back-substitution yields a unique solution for any a0, a1, a2. In other words, regardless of the values ofa0, a1, a2 we can always find coefficients c1, c2, c3 so that (∗) is true, that is the given vectors spanP2. And again, according to Theorem 19 (part (ii)), the vectors are a basis for P2.
81) Show that the set
S =
{(1 0−1 2
),
(−1 32 0
),
(1 1−1 0
),
(3 51 −1
)}is a basis for M22.
Solution:We do this problem in exactly the same way. Recall from notes that M22 has dimension equal to 4, andas the set S has exactly 4 vectors, according to Theorem 19 to show that we have a basis, all we needdo is show the given vectors are linearly independent (or, span M22).
For linearly independence we seek c1, c2, c3, c4 ∈ R such that
c1
(1 0−1 2
)+ c2
(−1 32 0
)+ c3
(1 1−1 0
)+ c4
(3 51 −1
)=
(0 00 0
),
i.e., after applying the usual operations for matrices (vector addition and scalar multiplication) we have(c1 − c2 + c3 + 3c4 3c2 + c3 + 5c4−c1 + 2c2 − c3 + c4 2c1 − c4
)=
(0 00 0
),
and so equating corresponding entries in both matrices yieldsc1 − c2 + c3 + 3c4 = 0
−c1 + 2c2 − c3 + c4 = 0
3c2 + c3 + 5c4 = 0
2c1 − c4 = 0
53
Then applying Gauss-Jordon to the associated coefficient matrix yields1 −1 1 3−1 2 −1 10 3 1 52 0 0 −1
−→1 0 0 00 1 0 00 0 1 00 0 0 1
= I4,
i.e., the linear system has the unique (‘trivial’) solution c1 = c2 = c3 = c4 = 0, i.e. the vectors arelinearly independent and hence according to Theorem 19 the given vectors are indeed a basis for M22.
Alternatively, if we decided to show instead that the vectors span M22, we go through almost the samesteps (exercise) leading to
c1 − c2 + c3 + 3c4 = a
−c1 + 2c2 − c3 + c4 = c
3c2 + c3 + 5c4 = b
2c1 − c4 = d
(with the assumption that a linear combination of the given vectors can be used to construct an arbitrary
vector
(a bc d
)∈M22). And as we saw with the linear independence argument the coefficient matrix
is non-singular, and so we obtain a unique solution for every right-hand-side vector (a, c, b, d)T . I.e.,the vectors do span M22, and hence according to Theorem 19 again we have that the given vectors area basis for M22.
82) Let
S =
123
, 2
0−4
, 0−37
,−41
6
(a) Without doing any calculations why do we know that S is not a basis for R3?(b) Show from first principles that S is not a basis for R3.
Hint: for (b) consider linear independence, but do not apply any row operations to the resulting matrixequation.
Solution:(a) S is not a basis for R3 because we are given 4 vectors when dimR3 = 3.
(b) Applying the usual argument for checking linear independence: we seek c1, c2, c3, c4 ∈ R such that
c1
123
+ c2
20−4
+ c3
0−37
+ c4
−416
=
000
, (?)
or (after using Theorem 1) 1 2 0 −42 0 −3 13 −4 7 6
c1c2c3c4
=
000
.
54
We see that the number of equations is < the number of unknowns =⇒ we have free variables =⇒a non-trivial solution =⇒ from (?) that the given vectors are linearly dependent, and so the vectorsare NOT a basis for R3.
Comment:However, the vectors do span R3. In fact the linear system resulting from the usual argument has aninfinite number of solutions, so there are an infinite number of ways of constructing an arbitrary vector(a, b, c)T ∈ R3 using the given vectors (Exercise).
4.7 Homogeneous Systems
83) Consider the homogeneous linear system Ax = 0 where
A =
1 2 −2 13 6 −5 41 2 0 3
.Find a basis for the null space of A (i.e., N(A)). What is the dimension of N(A)?
Solution:As shown in Algorithm 1, applying elementary row operations yields
A −→
1 2 0 30 0 1 10 0 0 0
(RREF).
With n = 4 unknowns and r = 2 pivot columns yields p = n− r = 2 free variables. (It’s good to dothis little check first so you know how many free variables we will get in the next step.) The associatedsystem of equations is {
x1 + 2x2 + 3x4 = 0
x3 + x4 = 0
The variables x4 and x2 are free. Set x4 = α, x2 = β so x3 = −α and x1 = −2β −−3α. So anysolution has the form
x =
x1
x2
x3
x4
=
−2β −−3α
β−αα
.Now to find the basis vectors that span the null-space of A observe that
x = α
−30−11
︸ ︷︷ ︸x1
+β
−2100
︸ ︷︷ ︸x2
.
So x1 and x2 clearly span the solution space, and as stated in the Algorithm they are also linearlyindependent. Thus a basis for N(A) is {x1, x2} and so the dimension of N(A) is 2.
55
4.8 Rank of a Matrix
84) Let
A =
(1 0 00 1 0
).
(a) Find expressions for the row space and the column space of A.(b) What are the dimensions of the row space and the column space of A?
Solution:(a) The row space of A is given by the set
span{(1, 0, 0), (0, 1, 0)} = {α(1, 0, 0) + β(0, 1, 0)︸ ︷︷ ︸=(α,β,0)
| α, β ∈ R}.
The column space of A is given by the set
span{(1
0
),
(01
),
(00
)}={α
(10
)+ β
(01
)+ γ
(00
)| α, β, γ ∈ R
}={(α
β
)| α, β ∈ R
}={α
(10
)+ β
(01
)| α, β ∈ R
}.
(b) The rows of A are linearly independent (exercise) and as they span the row space (by definition ofrow space) the rows of A are also a basis for the row space. The column space is spanned by (1, 0)T
and (0, 1)T (the vector (0, 0)T is redundant as it plays no part in the span of the columns). And as(1, 0)T and (0, 1)T are also linearly independent (exercise) they constitute a basis for the column spaceof A. So for both the row space and the column space the dimensions are equal, illustrating Theorem25.
85) Let
A =
(1 0 2 2 43 0 6 7 14
)(a) Find the dimension of the row space of A (i.e., the row rank of A).
(b) Find the dimension of the column space of A (i.e., the column rank of A).
Solution:(a) We use Theorem 24 and just count the number of non-zero rows in the RREF matrix. So applyingelementary row operations to A we get
A −→(1 0 2 0 00 0 0 1 2
)(RREF),
thus the dimension of the row space is 2.
(b) To find the column rank we apply elementary row operations to AT and count the number ofnon-zero rows in the RREF matrix:
AT −→
1 00 10 00 00 0
(RREF),
56
thus the dimension of the row space is 2.
Note:This problem illustrates Theorem 25.
86) Let
A =
(a bc d
),
for arbitrary a, b, c, d ∈ R. The matrix
B =
(a + c b + dc d
),
is obtained by applying the row operation r1 + r2 → r1 to A. Show that
row space of A = row space of B.
Solution:Row space of A:All entries have the form
α1(a, b) + α2(c, d) = (α1a + α2c, α1b + α2d).
Row space of B:All entries have the form
α1(a + c, b + d) + α2(c, d) = (α1(a + c) + α2c, α1(b + d) + α2d)
=(α1a + (α1 + α2)︸ ︷︷ ︸
α3
c, α1b + (α1 + α2)︸ ︷︷ ︸α3
d).
But the α’s are completely arbitrary, so the form of entries in the row space of A is the same form inthe row space of B. Thus A and B have the same row space.
Note:This problem can be generalized to prove Theorem 23.
87) Let
A =
1 −2 32 −5 11 −4 −7
Find a basis for the row space of A.
Solution:Applying elementary row operations yields (exercise)
A −→
1 0 130 1 50 0 0
(RREF)
Thus from Theorem 24 the nonzero rows of the matrix in RREF, i.e., (1, 0, 13) and (0, 1, 5) form abasis for the row space of A.
57
88) Find a basis for the subspace V of R3 spanned by S = {u1, u2, u3}, where
u1 =
1−23
, u2 =
2−51
, u3 =
1−4−7
(V = spanS),
(Hint: see Exercise 68 in the notes.)
Solution:Now we have that V is equal to the row space of the matrix
A =
1 −2 32 −5 11 −4 −7
.Thus from Exercise 87) the first and second rows of the matrix in RREF form a basis for the row spaceof A. Thus {v1, v2} where
v1 =
1013
, v1 =
015
is a basis for V = spanS.
89) Find the rank of the following matrices:
A =
1 3 −23 5 −10 1 −3
, B =
1 −2 30 1 −11 −1 2
.Solution:As we saw from Theorem 24 and Theorem 25 in order to find the rank of these matrices we reduce themto RREF and just count the number of non-zero rows:
A −→
1 0 00 1 00 0 1
(RREF),
(exercise) and so rankA = 3. Similarly
B −→
1 0 10 1 −10 0 0
(RREF),
(exercise) and so rankB = 2.
90) Consider the linear system Ax = b where
A =
1 2 0 −13 6 1 −11 2 −2 −5
, b =
01−1
.Consider the solvability of this system using the concept of rank.
Solution:Applying elementary row operations to the augmented matrix yields
[A|b] −→
1 2 0 −1 00 0 1 2 00 0 0 0 1
(RREF).
58
Thus we see using Theorem 24 that
2 = rankA 6= rank[A|b] = 3,
thus Theorem 28 tells us that the system is inconsistent.
Chapter 5
Inner Product Spaces
5.1 Length and Direction in Rn (n ≥ 2)
91) Consider the following vectors in R4:
u = (1,−1, 0, 4), v = (3, 2, 1, 0).
(a) Find the norms of u and v(b) Find the distance between u and v(c) Find the standard inner product of u and v(d) Find cos(θ) where θ is the angle between u and v(e) Find the normalized vector u corresponding to u(f) What value do we need to change the last component of v to in order for u and v to be orthogonal?
Solution:(a) The norm of u is
‖u‖ =√12 + (−1)2 + 02 + 42 =
√18 ≈ 4.24.
The norm of v is
‖v‖ =√32 + 22 + 12 + 02 =
√14 ≈ 3.74.
(b) The distance between u and v is
‖u− v‖ =√
(1− 3)2 + (−1− 2)2 + (0− 1)2 + (4− 0)2 =√30 ≈ 5.48.
(c) The (standard) inner product is given by
u · v = (1)(3) + (−1)(2) + (0)(1) + (4)(0) = 1.
(d)
cos(θ) =u · v‖u‖‖v‖
=1
√18√14
=1
3√2√14.
(e) All we do is divide each component of u by the norm of u (yielding a vector of unit length). Thusfrom (a):
u = (1/√18,−1/
√18, 0, 4/
√18).
59
(f) Let v = (3, 2, 1, x), where x is to be determined. For u and v to be orthogonal we need:
u · v = (1)(3) + (−1)(2) + (0)(1) + (4)(x) = 0,
i.e.3− 2 + 4x = 0 =⇒ 4x = −1 =⇒ x = −1/4.
5.2 Inner Product Spaces
92) Let u and v belong to an inner product space V . Given that ‖u‖ = 2, ‖v‖ = 3 and (u, v) = 4calculate ‖u + 2v‖2.
Solution:We use the axioms of an inner product listed in Definition 55 and the definition of an ‘induced’ norm inthe comments of page 239:
‖u + 2v‖2 = (u + 2v, u + 2v) (definition of the norm)
= (u, u + 2v) + (2v, u + 2v) (axiom (iii))
= (u + 2v, u) + (u + 2v, 2v) (axiom (ii))
= (u, u) + (2v, u) + (u, 2v) + (2v, 2v) (axiom (iii))
= (u, u) + (2v, u) + (2v, u) + 2(v, 2v) (axiom (ii) and (iv))
= (u, u) + 2(v, u) + 2(v, u) + 2(2v, v) (axiom (ii) and (iv))
= (u, u) + 4(u, v) + 4(v, v) (axiom (ii) and (iv))
= ‖u‖2 + 4(u, v) + 4‖v‖2 (definition of the norm)
= 22 + 4(4) + 4(3)2
= 56.
Note:The above argument is twice as long as it needs to be because of the explicit use of the ‘symmetry’axiom (ii). From now on we assume it applies, e.g. (u, cv) = c(u, v) (instead of (u, cv) = (cv, u) =c(v, u) = c(u, v)).
93) Consider the standard inner product (u, v) := u · v, where u, v ∈ R3.(a) Show that (u + v) · (u− v) = ‖u‖2 − ‖v‖2.(b) Verify the result in (a) for
u =
1−10
, v =
21−1
.Solution:(a)
(u + v) · (u− v) = u · u− u · v + v · u− v · v (Using Definition 55)
= ‖u‖2 − u · v + u · v − ‖v‖2 (Definition of the norm)
= ‖u‖2 − ‖v‖2 X
60
(b)
(u + v) · (u− v) =
1−10
+
21−1
· 1−10
− 2
1−1
=
30−1
·−1−2
1
= −3 + 0− 1
= −4.
And
‖u‖2 − ‖v‖2 =(12 + (−1)2 + 02
)−(22 + 12 + (−1)2
)= 2− 6
= −4 X
94) Let V = C[0, 1]∗ with the inner product
(f, g) :=
∫ 1
0f(x)g(x) dx,
and f(t) = 1 and g(x) = x.(a) Evaluate (f, g)(b) Without performing any calculations explain why (g, g) > 0(c) Define an induced norm for the inner product space V and hence calculate ‖f‖ and ‖g‖(d) Verify the Cauchy-Schwartz inequality |(f, g)| ≤ ‖f‖‖g‖Solution:(a)
(f, g) =
∫ 1
0x dx =
[x2
2
]10
=1
2.
(b)
(g, g) =
∫ 1
0x2 dx > 0,
as this is just the area under the graph of y = x2 from x = 0 to x = 1, which is positive.
(c) See the comment on page 239 of the Workbook:
‖f‖ =√
(f, f) =
√∫ 1
01 dx =
√[x]10 =
√1 = 1.
‖g‖ =√(g, g) =
√∫ 1
0x2 dx =
√[x3
3
]10
=
√1
3=
1√3.
(d) Using the results from (a) and (c) we have
0.5 =1
2= |(f, g)| ≤ ‖f‖‖g‖ = 1 ·
1√3≈ 0.5774 X
∗Set of continuous function on the interval [0, 1].
61
95) Complete Exercise 76 (page 243) in the Workbook by deriving formulae for a0 and am correspondingto Definition 57 for Fourier Series.(Hint: to derive the formulae for a0 and am, multiply (∗) (page 240 of the Workbook) by 1 andcos(mx) respectively, integrate from −π to π, and then use the Calculus results on page 241 of theWorkbook.)
Solution:
Derivation of the formula for a0:The formula (∗) is:
f(x) = a0 +∞∑n=1
(an cos(nx) + bn sin(nx)).
Following the hint we integrate both sides from −π to π yielding:
∫ π
−πf(x) dx = a0
∫ π
−π1 dx +
∞∑n=1
an∫ π
−πcos(nx) dx︸ ︷︷ ︸
=0
+bn
∫ π
−πsin(nx) dx︸ ︷︷ ︸
=0
= a0[x]
π−π = a0 · 2π,
thus a0 =1
2π
∫ π
−πf(x) dx.
Derivation of the formula for am:Following the hint (and using the Calculus results again) yields∫ π
−πf(x) cos(mx) dx = a0
∫ π
−πcos(mx) dx︸ ︷︷ ︸
=0
+∞∑n=1
{an
∫ π
−πcos(nx) cos(mx) dx
+ bn
∫ π
−πsin(nx) cos(mx) dx︸ ︷︷ ︸
=0
}
= am
∫ π
−πcos(mx) cos(mx) dx︸ ︷︷ ︸
=π
thus am =1
π
∫ π
−πf(x) cos(mx) dx.
Note on the 2nd to last line above that we used the fact that when n 6= m in the sum then∫ π−π cos(nx) cos(mx) dx = 0. We also can observe that when m = 0 the formula for am is consistent
with the formula for a0.
96) (a) Using Definition 57 in your Workbook find the Fourier coefficients and Fourier series of the ‘squarewave’ function
f(x) =
{0 if −π ≤ x < 0
1 if 0 ≤ x < πand f(x + 2π) = f(x),
i.e. f is periodic with period 2π and has the shape shown below (only shown on [−4π, 4π]):
(b) Graph several terms of your Fourier series against the square wave to see how well successive numbersof terms approximates the given function.
62
Solution:(a) Finding the Fourier coefficients:
a0 =1
2π
∫ π
−πf(x) dx =
1
2π
∫ 0
−πf(x) dx +
1
2π
∫ π
0f(x) dx
=1
2π
∫ 0
−π0 dx +
1
2π
∫ π
01 dx
= 0 +1
2π(π)
=1
2.
an =1
π
∫ π
−πf(x) cos(nx) dx =
1
π
∫ 0
−πf(x) cos(nx) dx +
1
π
∫ π
0f(x) cos(nx) dx
=1
π
∫ 0
−π0 · cos(nx) dx +
1
π
∫ π
01 · cos(nx) dx
= 0 +1
π
[sin(nx)
n
]π0
=1
nπ(sin(nπ)− sin(0))
= 0.
bn =1
π
∫ π
−πf(x) sin(nx) dx =
1
π
∫ 0
−πf(x) sin(nx) dx +
1
π
∫ π
0f(x) sin(nx) dx
=1
π
∫ 0
−π0 · sin(nx) dx +
1
π
∫ π
01 · sin(nx) dx
= 0 +1
π
[−
cos(nx)
n
]π0
= −1
nπ(cos(nπ)− cos(0)) (∗)
Now
cos(0) = 1, cos(π) = −1,cos(2π) = 1, cos(3π) = −1, etc.
63
Thus from (∗) we have (recall, n ≥ 1)
bn =
{− 1nπ
(+1− 1) n even
− 1nπ
(−1− 1) n odd=
{0 n even2nπ
n odd.
Finding the Fourier Series:We have
f(x) = a0 + a1 cos(x) + b1 sin(x)
+ a2 cos(2x) + b2 sin(2x)
+ a3 cos(3x) + b3 sin(3x) + . . .
Now we saw that all terms that involve cosine, and sine terms with even multiples of x in the argumentare zero. So we have
f(x) =1
2+ b1 sin(x) + b3 sin(3x) + b5 sin(5x) + . . .
=1
2+
2
πsin(x) +
2
3πsin(3x) +
2
5πsin(5x) + . . .
(b) It is natural to wonder how well the series approximates the square wave when only a few terms areused. In the graphs below† we denote
Sn =1
2+
2
πsin(x) +
2
3πsin(3x) + · · · +
2
nπsin(nx).
Even though the square wave is discontinuous and the partial sums Sn are continuous, the more termsin the Fourier series we use the more closely we approximate the square wave.
97) Let u and v be vectors in an inner product space V. Prove that
‖u + v‖ ≤ ‖u‖ + ‖v‖
with equality if and only if u and v are orthogonal.
†Reproduced from ‘Stewart: Calculus, Sixth Edition.
64
(Hint: start by expanding ‖u + v‖2 with the aid of the rules in Definition 55, use the definition of aninduced norm (see page 239 in the Workbook), and then use the Cauchy-Schwartz inequality.)
Solution:
‖u + v‖2 = (u + v, u + v)
= (u + v, u) + (u + v, v)
= (u, u) + (v, u) + (u, v) + (v, v)
= ‖u‖2 + 2(u, v) + ‖v‖2 (∗)
Using the Cauchy-Schwartz inequality (see Theorem 33) we have
‖u + v‖2 ≤ ‖u‖2 + 2‖u‖‖v‖ + ‖v‖2 = (‖u‖ + ‖v‖)2,
and so taking the square root of both sides yields
‖u + v‖ ≤ ‖u‖ + ‖v‖ X
We also observe from (∗) that if u and v are orthogonal (i.e. (u, v) = 0) then we get the equality case:‖u + v‖ = ‖u‖ + ‖v‖.
Chapter 6
Eigenvalues and Eigenvectors
6.1 Eigenvalues and Eigenvectors
98) Prove Theorem 35 in the Workbook which states: if A is a square n×n matrix and λ is an eigenvalueof A, then the eigenspace of λ is a subspace of Rn.
Solution:The eigenspace of λ is given by
V = {x ∈ Rn | Ax = λx}.
So x1, x2 ∈ V then Ax1 = λx1 and Ax2 = λx2. Now consider
A(x1 + x2) = Ax1 +Ax2 = λx1 + λx2 = λ(x1 + x2),
hence x1 + x2 ∈ V , i.e., V is closed under vector addition. Also consider x ∈ V and k ∈ R, then
A(kx) = kAx = kλx = λ(kx),
hence kx ∈ V , i.e., V is closed under scalar multiplication. Thus according to Theorem 15 V is asubspace of Rn.
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A note about notation:Notice how I didn’t underline vectors in this problem, i.e. I used x instead of x. Now most advancedtextbooks don’t bother with the underlying as the context should make it clear when we are dealingwith a vector and when we are dealing with a scalar.∗ However, for the standard eigenvector-eigenvectorproblems (see next exercise) we really do need to distinguish between a vector and its components, so Ido underline vectors.
99) Find the eigenvalues and eigenvectors of the matrix
A =
(2 −121 −5
).
Solution:
We seek the solution ofAx = λx, i.e. find the eigenvalues λ and the associated eigenvectors x =
(x1
x2
)such that (
2 −121 −5
)(x1
x2
)= λ
(x1
x2
)I.e. {
2x1 − 12x2 = λx1
x1 − 5x2 = λx2
or, {(2− λ)x1 − 12x2 = 0
x1 + (−5− λ)x2 = 0
which in matrix form is((2− λ) −12
1 (−5− λ)
)(x1
x2
)=
(00
)(∗)
We seek the non-trivial solutions (x 6= 0) so we know from lecture notes that we need
|A− λI| =∣∣∣∣(2− λ) −12
1 (−5− λ)
∣∣∣∣ = 0,
i.e.
(2− λ)(−5− λ) + 12 = 0
=⇒ −10− 2λ + 5λ + λ2 + 12 = 0
=⇒ λ2 + 3λ + 2 = 0
=⇒ (λ + 1)(λ + 2) = 0.
Hence we have the eigenvalues λ1 = −1 and λ2 = −2. To find the associated eigenvectors we solve(∗) for x for each value of λ.
∗It also makes it quicker for me to type up the solutions :).
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Case λ1 = −1:From (∗) we have (
3 −121 −4
)(x1
x2
)=
(00
).
Elementary row operations yield (exercise)(3 −121 −4
)−→
(1 −40 0
),
thus x1 − 4x2 = 0, or x1 = 4x2 where x2 is free. Set x2 = α so the eigenvectors associated with theeigenvalue λ1 have the form
x1 =
(x1
x2
)=
(4αα
)= α
(41
).
E.g., for α = 1 we get the particular eigenvector
(41
).
Case λ2 = −2:From (∗) we have (
4 −121 −3
)(x1
x2
)=
(00
).
Elementary row operations yield (exercise)(4 −121 −3
)−→
(1 −30 0
),
thus x1 − 3x2 = 0, or x1 = 3x2 where x2 is free. Set x2 = β so the eigenvectors associated with theeigenvalue λ2 have the form
x2 =
(x1
x2
)=
(3ββ
)= β
(31
).
E.g., for β = 2 we get the particular eigenvector
(62
).
6.2 Diagonalization and Similar Matrices
100) Diagonalize the matrix A from Exercise 99). Justify this process.
Solution:We saw in Exercise 99) that the eigenvectors of A were distinct (i.e., different), thus we know from
Corollary 6 of Theorem 38 that A is indeed diagonalizable. We had A =
(2 −121 −5
)with λ1 = −1
and λ2 = −2. Corresponding to λ1 and λ2 we found the particular eigenvalues
x1 =
(41
)and x2 =
(62
).
(Note: to diagonalize A it doesn’t matter which particular eigenvectors you use.) We let
P = [x1|x2] =
(4 61 2
)and D =
(λ1 00 λ2
)=
(−1 00 −2
).
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Let’s check our answers. First observe that |P | = 8 − 6 = 2 6= 0 so P is nonsingular. Then usingTheorem 37
A = PDP−1 =
(4 61 2
)(−1 00 −2
)1
2
(2 −6−1 4
)=
(2 −121 −5
)X.
101) Use diagonalization of the matrix A in Exercise 99) to calculate A10.
Solution:Recall from Exercise 100) we found that
A = PDP−1 =
(4 61 2
)(−1 00 −2
)1
2
(2 −6−1 4
).
Thus using Theorem 39 we have
A10 = PD10P−1 =
(4 61 2
)((−1)10 0
0 (−2)10)
1
2
(2 −6−1 4
)=
(4 61 2
)(1 00 1024
)1
2
(2 −6−1 4
)=
(−3068 12276−1023 4093
).
Note:This is easily checked in Matlab.
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