Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 127
As a quick check, we can take a vector at random
(
4
⃗v =
5)
giving
( ) ( ( ) (
4 1 1 4 9
Rep E2
(⃗v) =
= = t(⃗v)
5 3 −1)
5 7)
while the calculation with respect
(
to B,
)
B
( ) ( ( )
1 −2 0 1 −2
Rep B (⃗v) =
=
−3 −5 2 −3)
−11
B,B B
B
yields the same result.
( ( ) (
1 −1 9
−2 · − 11 · =
2)
−1 7)
(b) We have
R 2 t
w.r.t. E 2
−−−−→ R 2 w.r.t. E T 2
⏐
⏐
id↓
id↓
Rep B,B (t) = Rep E2 ,B(id) · T · Rep B,E2 (id)
R 2 t
w.r.t. B −−−−→ R 2 w.r.t. B
ˆT
and, as in the first item of this question
( )
Rep B,E2 (id) = ⃗β1 · · · βn ⃗
Rep E2,B(id) = Rep B,E2 (id) −1
so, writing Q for the matrix whose columns are the basis vectors, we have that Rep B,B (t) = Q −1 T Q.
Three.V.2.25
(a) The adapted form of the arrow diagram is this.
h
V w.r.t. B1 −−−−→ W w.r.t. D
H
⏐
⏐
id↓Q
id↓P
h
V w.r.t. B2 −−−−→ W w.r.t. D
Ĥ
Since there is no need to change bases in W (or we can say that the change of basis matrix P is the
identity), we have Rep B2 ,D(h) = Rep B1 ,D(h) · Q where Q = Rep B2 ,B 1
(id).
(b) Here, this is the arrow diagram.
h
V w.r.t. B −−−−→ W w.r.t. D1
H
⏐
⏐
id↓Q
id↓P
h
V w.r.t. B −−−−→ W w.r.t. D2
Ĥ
We have that Rep B,D2 (h) = P · Rep B,D1 (h) where P = Rep D1 ,D 2
(id).
Three.V.2.26
(a) Here is the arrow diagram, and a version of that diagram for inverse functions.
V w.r.t. B
⏐
id
↓Q
V w.r.t.
ˆB
h
−−−−→
H
h
−−−−→
Ĥ
W w.r.t. D
⏐
id
↓P
W w.r.t.
ˆD
V w.r.t. B
⏐
id
↓Q
V w.r.t.
ˆB
h −1
←−−−−
H −1
W w.r.t. D
⏐
id
↓P
h −1
←−−−−
Ĥ −1 W w.r.t. ˆD
Yes, the inverses of the matrices represent the inverses of the maps. That is, we can move from the
lower right to the lower left by moving up, then left, then down. In other words, where Ĥ = P HQ
(and P, Q invertible) and H, Ĥ are invertible then Ĥ−1 = Q −1 H −1 P −1 .
(b) Yes; this is the prior part repeated in different terms.
(c) No, we need another assumption: if H represents h with respect to the same starting as ending
bases B, B, for some B then H 2 represents h ◦ h. As a specific example, these two matrices are both
rank one and so they are matrix equivalent
( 1
) 0
( 0
) 0
0 0 1 0
but the squares are not matrix equivalent — the square of the first has rank one while the square of
the second has rank zero.