121
Let V1 ≡ span
1
10
,
101
,V2 ≡ span
0−12
. Show thatR3 =V1⊕V2 and
and that Vi is A invariant. Find the matrix of A with respect to the ordered basis 1
10
,
101
,
0−12
(*)
First note that 1 0 01 0 −1−2 2 3
1
10
=
110
,
1 0 01 0 −1−2 2 3
1
01
=
101
Therefore, A(V1)⊆V1. Similarly, 1 0 0
1 0 −1−2 2 3
0−12
=
0−24
= 2
0−12
and so A(V2) ⊆ V2. The vectors in ∗ clearly are a basis for R3. You can verify this byobserving that there is a unique solution x,y,z to the system of equations
x
110
+ y
101
+ z
0−12
=
abc
for any choice of the right side. Therefore, by Lemma 6.0.3, R3 =V1⊕V2.
If you look at the restriction of A to V1, what is the matrix of this restriction? It satisfiesA
110
,A
101
=
1
10
,
101
( a b
c d
)
Thus, from what was observed above, you need the matrix on the right to satisfy 1
10
,
101
=
1
10
,
101
( a b
c d
)
and so the matix on the right is just
(1 00 1
). As to the matrix of A restricted to V2, we
need
A
0−12
= 2
0−12
= a
0−12