Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 125
⎛
( )
1 0 0
(a)
(b) ⎝ 1 0 0 0
⎞
0 1 0 0⎠
0 0 0
0 0 1 0
Three.V.2.12 Recall the diagram and the formula.
R 2 t
w.r.t. B −−−−→ R 2 w.r.t. D
T
⏐
⏐
id↓
id↓
ˆT = Rep D, ˆD(id) · T · Rep ˆB,B
(id)
(a) These two
show that
R 2 w.r.t.
ˆB
(
1
= 1 ·
1)
t
−−−−→
ˆT
and similarly these two (0 )
1
give the other nonsinguar matrix.
R 2 w.r.t.
ˆD
( ) ( ) (
−1 2 1
+ 1 ·
= (−3) ·
0 1 −1)
Rep D, ˆD(id) =
( )
1 −3
1 −1
( ) (
−1
2
+ (−1) ·
0
1)
( ( ) ( ( (
1 0 1 1 0
= 0 · + 1 ·
= 1 · + 1 ·
0)
1 1)
0)
1)
Rep ˆB,B
(id) =
( )
0 1
1 1
Then the answer is this. ( ) ( ) ( ) ( )
1 −3 1 2 0 1 −10 −18
ˆT = =
1 −1 3 4 1 1 −2 −4
Although not strictly necessary, a check is reassuring. Arbitrarily fixing
(
3
⃗v =
2)
we have that
and so t(⃗v) is this.
(
3
Rep B (⃗v) =
2)
B
( ) (
1 2 3
=
3 4 2)
B,D B
( ( ( )
1 1 24
7 · + 17 · =
1)
−1)
−10
Doing the calculation with respect to ˆB, ˆD starts with
( ) ( )
−1 −10 −18
Rep ˆB(⃗v) =
3 −2 −4
and then checks that this is the same result. ( ) (
−1 2
−44 · − 10 · =
0 1)
(b) These two (
1
1)
show that
and these two (
1
2)
show this.
ˆB
ˆB, ˆD
( )
−1
=
3 ˆB
( )
24
−10
( )
7
17
D
( )
−44
−10 ˆD
= 1 (
1
3 2)
· + 1 ( ) ( ( (
2 1 1 2
3 ·
= −1 · + 1 ·
1 −1)
2)
1)
Rep D, ˆD(id) =
(
1/3
)
−1
1/3 1
( ( ) ( ( (
1 0 1 1 0
= 1 · + 2 ·
= −1 · + 0 ·
0)
1 0)
0)
1)
Rep ˆB,B
(id) =
( ) 1 1
2 0
With those, the conversion goes in this way.
( ) ( ) ( )
1/3 −1 1 2 1 1
ˆT = =
1/3 1 3 4 2 0
(
−28/3
)
−8/3
38/3 10/3