MH1201 - Solution Set 9 - MH1201 - SOLUTIONS TO PROBLEM SET...

MH1201 - SOLUTIONS TO PROBLEM SET 9
Problem 1 .
(1) We have















T ( β
1
) = T


1
2


=


- 2
- 3


= ( - 1) β
2
= 0 β
1
+ ( - 1) β
2
T ( β
2
) = T


2
3

 =


2
4

 = 2 β
1
= 2 β
1
+ 0 β
2
,
and hence
[ T ]
β
=
"
0 2
- 1 0
#
.
(2) It is clear, by 1), that the basis β does not consist of eigenvectors of T .
Problem 2 .
It’s similar to Problem 1.
(1) We have













T ( β
1
) = T (1 - x + x 3 ) = - 1 + x - x 3 = - β
1
= ( - 1) β
1
+ 0 β
2
+ 0 β
3
+ 0 β
4
T ( β
2
1) = T (1 + x
2
) = - x - x
2
+ x
3
= β
1
- β
2
= 1 β
1
+ ( - 1) β
2
+ 0 β
3
+ 0 β
4
T ( β
3
) = T (1) = x 2 = β
2
- β
3
= 0 β
1
+ 1 β
2
+ ( - 1) β
3
+ 0 β
4
T ( β
4
) = T ( x + x
2
) = - x - x
2
= - β
4
= 0 β
1
+ 0 β
2
+ 0 β
3
+ ( - 1) β
4
,
and hence
[ T ]
β
=





- 1 1 0 0
0 - 1 1 0
0 0 - 1 0
0 0 0 - 1





.
(2) It is clear, by 1), that the basis β does not consist of eigenvectors of T .
Problem 3 .
It’s similar to Problem 4, Tutorial # 8 last week. We have A =
"
1 2
3 2
#
.
1

2