Solutions to Assignment 6 1. If L is a linear transformation from V to ...

Solutions to Assignment 6
1. If L is a linear transformation from V to W , use mathematical induction to prove
that
Proof:
L(c1v1 + c2v2 + · · · + cnvn) = c1L(v1) + c2L(v2) + · · · + cnL(vn)
When n = 1, L(c1v1) = c1L(v1) by the definition of linear transformations.
Assume the formula is true for n = k, i. e.,
L(c1v1 + c2v2 + · · · + ckvk) = c1L(v1) + c2L(v2) + · · · + ckL(vk).
Consider the case n = k + 1:
L(c1v1 + c2v2 + · · · + ckvk+1)
= L((c1v1 + c2v2 + · · · + ckvk) + ck+1vk+1) [(c1v1 + · · · + ckvk) is a vector]
= L(c1v1 + c2v2 + · · · + ckvk) + ck+1L(vk+1) [Because L is linear]
= c1L(v1) + c2L(v2) + · · · + ckL(vk) + ck+1L(vk+1)
[Formula is assumed to be true for n = k]
So the formula also holds true for n = k + 1. Then by mathematical induction on
n, L(c1v1 + c2v2 + · · · + cnvn) = c1L(v1) + c2L(v2) + · · · + cnL(vn) is true for every
positive integer n provided that L is linear.
2. Let E =
1 2
, and F =
1 3
2 1
, be ordered bases for R
1 0
2 , and suppose
L : R2 → R2 is defined by L(x) = (8x1 − 6x2, 9x1 − 7x2) T .
(a) Find the standard matrix representation of L.
(b) Find the the matrix representing L with respect to the bases E (in the domain)
and F (in the target space).
(c) Find the matrix representing L with respect to the basis E in both the domain
and the target space.
(d) Use part (c) to find the matrix representing L k with respect to the basis E
in both the domain and the target space.
Answer:

(a) The standard matrix representation is
A = L
1
, L
0
0
=
1
(b) The matrix for L relative to bases E, F is
B = F −1
L
1
, L
1
2
3
= 1
−1
(c) The matrix for L relative to basis E is
C = E −1
L
1
, L
1
2
3
Alternatively,
C = E −1 AE =
= 1
1
0 −1
−1 2
3 −2
−1 1
8 −6
9 −7
3
−2 8
−6 1
2
−1 1 9 −7 1 3
(d) The matrix for L k relative to basis E is C k =
2 −2
2 −3
2 −2
2 −3
=
=
=
2 0
0 −1
2
−3
−2 4
2 0
0 −1
k
k 2 0 2 0
=
0 −1 0 (−1) k
.
3. Let A and B be similar matrices and let λ be any scalar. Show that:
(a) det(A) = det(B).
(b) A − λI and B − λI are similar.
Answer:
(a) A = S −1 BS for some nonsingular matrix S. Then
det(A) = det(S −1 BS) = det(S −1 ) det(B) det(S) =
(b) Note that λI = S −1 (λI)S for any scalar λ. Then
A − λI = S −1 BS − S −1 (λI)S = S −1 (B − λI)S.
1
det(B) det(S) = det(B).
det(S)
So by definition of similar matrices, A − λI and B − λI are similar for any
scalar λ.
2