Linear Algebra Math 221 Open Book Exam 3 Open Notes 21 Nov, 2002 Calculators Permitted Show all work (except #4)

1. (30 pts) Given the matrix

−=110

240

001

A

Note throughout this problem that the matrix is block diagonal, with a 1x1 matrix (1) and a 2x2 matrix (4, -2; 1, 1). Each of these problems can be solved separately for the two submatrices, and then combined for the final answer. The characteristic polynomial of matrix A is the product of the char polynomials of the two submatrices, char(A) = char(1)char(4,-2;1,1) = (1-λ)[(1-λ)(1-λ)-(-2)(1)]. The eigenvalues of A are the union of the eigenvalues of the two submatrices. Thus eigenvals(A) = eigenvals(1) U eigenvals(4,-2;1,1) = {1}U{2,3} = {1,2,3}. The eigenvectors for A are the eigenvectors of the submatrices, with suitable padding.

a) (5 pts) Compute the characteristic polynomial of A.

char(A) = det (A-λI) =

−−−

−=

−

−

λλ

λλ

110

240

001

det

100

010

001

110

240

001

det

= (1-λ)(4-λ)(1-λ)+(0)(-2)(0)+(0)(1)(0)–(0)(4-λ)(0)–(0)(0)(1-λ)–(-2)(1)(1-λ) = (1-λ)[(4-λ)(1-λ)-(-2)(1)] = (1-λ)[λ2-5λ+6] = -λ3+6λ2-11λ+6 Answer: char(A) = -λ3+6λ2-11λ+6 Check: The constant term of char(A) is equal to det(A) = 6.

b) (10 pts) Compute the eigenvalues of A.

The eigenvalues of A are the roots of the characteristic polynomial. We note that the characteristic polynomial has the form (1-λ)[λ2-5λ+6], so it has λ=1 as one root. The other two roots can be found as the roots of the quadratic polynomial λ2-5λ+6. Using the quadratic formula we get roots

{ }3,22

15)1(2

)6)(1(4)5()5( 2

=±

=−−±−−

=λ

Answer: λ = {1, 2, 3} Check: The product of the eigenvalues is the determinant.

c) (10 pts) Find the eigenvectors of A.

We compute the eigenvectors for a given eigenvalue λ by solving the matrix equation (A-λI) = 0 and selecting a basis for the solution space. λ = 1:

(A-λI) = (A-I) =

−→

−→

−

000

200

010

000

230

010

010

230

000

So the solution is:

==

0

0

free is

3

2

1

x

x

x

so

=

=

0

0

1

0

0 1

1

3

2

1

x

x

x

x

x

So the eigenvector corresponding to the eigenvalue λ=1 is the vector (1;0;0), or any multiple of it. Check: Confirm that A(1;0;0) = (1;0;0)

λ = 2:

(A-λI) = (A-2I) =

−

−→

−

−→

−−

−

000

110

001

000

220

001

110

220

001

So the solution is:

==

free is

0

3

32

1

x

xx

x

so

=

=

1

1

00

3

3

3

3

2

1

x

x

x

x

x

x

So the eigenvector corresponding to the eigenvalue λ=2 is the vector (0;1;1), or any multiple of it. Check: Confirm that A(0;1;1) = 2(0;1;1) = (0;2;2)

λ = 3:

(A-λI) = (A-I) =

−

−→

−−

−

000

210

002

210

210

002

So the solution is:

==

free is

2

0

3

32

1

x

xx

x

so

=

=

1

2

0

2

0

3

3

3

3

2

1

x

x

x

x

x

x

So the eigenvector corresponding to the eigenvalue λ=3 is the vector (0;2;1), or any multiple of it. Check: Confirm that A(0;2;1) = 3(0;2;1) = (0;6;3)

Answer: λ = 1, 2, 3 have eigenvectors

1

2

0

,

1

1

0

,

0

0

1

respectively

d) (5 pts) Find a matrix P such that P-1AP is diagonal.

The matrix P consists of the linearly independent eigenvectors of A.

Answer: P =

110

210

001

Check: P-1AP =

=

−

−

300

020

001

110

210

001

110

240

001

110

210

0011

= D

Note that if we put the eigenvectors together to form P in a different order we will still get a diagonal matrix D, but the order of the eigenvalues along the diagonal will be different.

2. (25 pts) If

=

233

242

112

121

A

a) (10 pts) Compute rank(A)

We compute rank(A) by row reducing A and counting the number of pivots (equivalently, the number of non-zero rows).

−−

→

−−−−

→

−−

−−→

000

000

130

121

000

130

130

121

130

000

130

121

233

242

112

121

Answer: Thus rank(A) = 2

b) (5 pts) Find a basis for the column space of A.

We compute a basis for the column space by selecting those columns in the original matrix A corresponding to the pivot columns in the reduced form of A. As there are pivots in columns 1 and 2 we select the first and second columns of A as a basis of the column space. (As it happens, for this value of A we could have selected any two columns, but that isn’t always the case.)

Answer: A basis of col(A) is

3

4

1

2

,

3

2

2

1

c) (10 pts) Find a basis for the null space of A.

We compute a basis for the null space by solving the homogeneous matrix equation Ax=0 and finding a basis of the solutions. From (a) above:

→

→

−−

→

000

000

10

01

000

000

10

121

000

000

130

121

233

242

112

121

3

13

1

3

1

so

−=−=

free is

3/

3/

3

32

31

x

xx

xx

and our solutions have the form

−−

=

−−

=

13

13

1

3

3

3

3

3

2

1

3

3

x

xx

x

xx

x

. Thus

−−

13

13

1

, or any

multiple of it, is a basis for the null space.

Answer: A basis of Null(A) is

− 3

1

1

Check: Confirm that Av = 0, for each vector v in the basis of Null(A)

3. (20 pts) Prove that if A is an invertible matrix and {v1, v2, v3, v4} are linearly

independent, then {Av1, Av2, Av3, Av4} are linearly independent. Proof: Assume not – in other words assume that {Av1, Av2, Av3, Av4} is linearly dependent. Thus, for some non-zero coefficients ai we have: a1(Av1)+ a2(Av2)+ a3(Av3)+ a4(Av4)=0 defn of linearly depn A(a1v1)+ A(a2v2)+ A(a3v3)+ A(a4v4)=0 Acx = cAx [Thm 5, Chap 1] A(a1v1+a2v2+a3v3+a4v4) = 0 A(u+v) = Au + Av [Thm 5, Chap 1] A-1A(a1v1+a2v2+a3v3+a4v4) = A-10 A is invertible a1v1+a2v2+a3v3+a4v4 = 0 But this is a contradiction, as we know that {v1, v2, v3, v4} are linearly independent. So our assumption that {Av1, Av2, Av3, Av4} are linearly dependent is false, and they must be linearly independent.

4. (25 pts) Multiple Choice: ALWAYS/SOMETIMES/NEVER (5 pts per question)

a) A 3x3 matrix with 2 distinct eigenvalues is diagonalizable.

SOMETIMES – If the repeated eigenvalue has two linearly independent eigenvectors then yes, if not then no. Examples of the two cases are

200

020

001

diagonalizable, and

200

120

001

non-diagonalizable. [Chap 5,

Suppl Problems, #1o]

b) The rank of a 3x3 invertible matrix is 1.

NEVER – An invertible matrix has full rank (the same as the number of rows or columns). See the continuation of the Invertible Matrix Thm, Sect 4.6, pg 267.

c) If A is row equivalent to B, then A and B have the same eigenvalues.

SOMETIMES – In general no, but there are some (rare) cases for which it is true. Do not confuse row equivalence with similarity (i.e. P-1AP), which does preserve eigenvalues and determinant. Also do not confuse this with the fact that the replacement row operation does preserve determinant (but not eigenvalues).

d) A 3x3 matrix with 3 distinct eigenvalues is diagonalizable.

ALWAYS – Each eigenvalue has at least one eigenvector, and eigenvectors from different eigenvalues are linearly independent. Thus, three distinct eigenvalues gives us three linearly independent eigenvectors, and we can construct P and diagonalize the matrix. [Essentially 1999 Exam 3, problem #4e]

e) If a vector is in the Null space of A then it is also in the Null space of BA.

ALWAYS – If the vector x is in Null(A), then Ax=0. Thus B(Ax) = B(0) = 0 and we see that (BA)x = 0 and x is in Null(BA). [2001 Exam 3, problem #1]