Math 415 - Fall 2011 - Exam 3 - SolutionsProblem 1 (3 points). Indicate whether the following statements are true or false.

(1) T F There is an isomorphism between M2×3(R) and R5.(2) T F The linear transformation T : P3(R)→ P2(R) defined by T ( f ) = f ′ is onto.(3) T F Any 3 orthonormal vectors in R3 form a basis for R3.(4) T F If det(A) = det(B) then A and B are similar matrices.(5) T F If A is a square matrix, then det(AAT ) is always non-negative.(6) T F If V and W are finite dimensional vector spaces with dim(V ) = dim(W ), then they are isomorphic.

Answer:1) False, they have different dimensions, see Theorem 4.2.4(b).2) True, for any g := a+bx+ cx2 ∈ P2(R), take f := ax+ 1

2bx2 + 13cx3 ∈ P3(R). Then T ( f ) = g.

3) True, see Theorem 5.1.34) False, for example let A be the zero matrix and B be any nonzero matrix with determinant zero. Then A andB are not similar, but have equal determinants.5) True, since det(AAT ) = det(A)det(AT ) = det(A)det(A) = det(A)2 ≥ 06) True, if n := dim(V ) = dim(W ), then both V and W are isomorphic to Rn (see for instance Theorem 4.2.3)so they are isomorphic to each other.

Problem 2 (1 point). Which of the following maps from M2×2(R) to M2×2(R) is linear?a) T (M) = M+ I2, b) T (M) = MT , c) T (M) = det(M)I2, d) T (M) = M2.Answer: (b)(a) is not linear since T (0) 6= 0.(c) is not linear since det(M+M′) 6= det(M)+det(M′) in general.(d) is not linear since (M+M′)2 6= (M)2 +(M′)2 in general.

Problem 3 (1 point). Let A be a fixed invertible 2 by 2 matrix. Which of the following maps from M2×2(R) toM2×2(R) is an isomorphism?a) T (M) = M+A, b) T (M) = AM−MA, c) T (M) = det(M)A, d) T (M) = AMA−1.Answer: (d), see for instance example 5 on page 168.(a) is not even linear since T (0) 6= 0.(b) is linear, but not an isomorphism since T (A) = AA−AA = 0, so kernel of T is not zero.(c) is not even linear since det(M+M′) 6= det(M)+det(M′) in general.

Problem 4 (2 points). Let M =

k 1 0 00 0 0 k0 0 k 0

3k−2 k 0 0

.

i) Compute the determinant of M.Answer: Using the permutation definition of determinant, det(M) =−k4 + k2(3k−2) =−k2(k−1)(k−2)ii) Determine for which values of the constant k the matrix M is invertible.Answer: M is invertible ⇐⇒ det(M) 6= 0 ⇐⇒ k 6= 0,1,2.

1

2

Problem 5 (2 points). Let A =

[a 11 b

]and B =

[ √2 −1−1

√2

].

i) Compute det(A−3)Answer: det(A−3) = det(A)−3 = (ab−1)−3 = 1/(ab−1)3

ii) Compute det(AB5A).Answer: det(AB5A) = det(A)det(B)5det(A) = (ab−1)(1)5(ab−1) = (ab−1)2

Problem 6 (3 points). Let T be the linear transformation fro T : P2(R) to P2(R) defined by

T ( f ) = f − (x+1) f ′

i) Compute the matrix representation of T with resect to the standard basis of P2(R).ii) Give a basis for the kernel of T .iii) Give a basis for the image of T .

Answer: Note that the standard basis of P2(R) is β = {1,x,x2} and the order matters.i) First we computeT (1) = 1− (x+1)(0) = 1 = 1 ·1+0 · x+0 · x2

T (x) = x− (x+1)(1) =−1 =−1 ·1+0 · x+0 · x2

T (x2) = x2− (x+1)(2x) =−2x− x2 = 0 ·1−2 · x−1 · x2

Hence [T ]β =[[T (1)]β|[T (x)]β|[T (x2)]β

]=

1 −1 00 0 −20 0 −1

.

ii) First compute ker([T ]β) = Span

1

10

. Hence ker(T ) = Span{1+ x}.

iii) First compute im([T ]β) = column space of [T ]β = Span

1

00

, 0−2−1

.

Hence im(T ) = Span{1,−2x− x2}.

Problem 7 (3 points). Consider the basis β for R2 consisting of vectors v1 =

[−3

4

]and v2 =

[−1−7

].

i) Use Gram-Schmidt process to convert β to an orthonormal basis γ = {u1,u2} for R2.ii) Compute the change of coordinate matrix R from β to γ.

iii) Compute the QR factorization of the matrix M =

[−3 −1

4 −7

]Answer:i) As in the Gram-Schmidt process, u1 =

v1||v1|| =

v15 =

[−3

545

].

Compute v2 ·u1 =−5, hence v2− (v2 ·u1)u1 = v2 +5u1 =

[−4−3

]Thus, u2 =

v2−Pro ju1(v2)

||v2−Pro ju1(v2)|| =v2−(v2·u1)u1||v2−(v2·u1)u1|| =

15

[−4−3

]=

[−4

5−3

5

].

ii) R =[[v1]γ|[v2]γ

]=

[v1 ·u1 v2 ·u1v1 ·u2 v2 ·u2

]=

[5 −50 5

].

3

iii) Let Q = [u1|u2] =

[−3

5 −45

45 −3

5

]. Then the QR factorization of M = [v1|v2] is

M = Q ·R = [u1|u2] ·R =

[−3

5 −45

45 −3

5

]·[

5 −50 5

].