Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 111
Three.IV.2.29 The ‘commutes with’ relation is reflexive and symmetric. However, it is not transitive:
for instance, with ( ) ( ) ( )
1 2 1 0 5 6
G = H = J =
3 4 0 1 7 8
G commutes with H and H commutes with J, but G does not commute with J.
Three.IV.2.30
(a) Either
⎛ ⎞
of these.
⎛ ⎞
⎛
⎞
⎝ x y⎠ ↦−→
πx ⎝ x 0⎠ y
↦−→ ⎝ 0 0⎠
⎝ x y⎠ y
↦−→ ⎝ 0 y⎠ ↦−→
πx ⎝ 0 0⎠
z 0 0 z 0 0
(b) The composition is the fifth derivative map d 5 /dx 5 on the space of fourth-degree polynomials.
(c) With respect to the natural bases,
⎛ ⎞
⎛ ⎞
1 0 0
0 0 0
Rep E3,E 3
(π x ) = ⎝0 0 0⎠ Rep E3,E 3
(π y ) = ⎝0 1 0⎠
0 0 0
0 0 0
and their product (in either order) is the zero matrix.
(d) Where B = 〈1, x, x 2 , x 3 ,
⎛
x 4 〉,
⎞
⎛
⎞
0 0 2 0 0
0 0 0 6 0
Rep B,B ( d2
dx 2 ) = 0 0 0 6 0
⎜0 0 0 0 12
⎟ Rep B,B ( d3
⎝0 0 0 0 0 ⎠
dx 3 ) = 0 0 0 0 24
⎜0 0 0 0 0
⎟
⎝0 0 0 0 0 ⎠
0 0 0 0 0
0 0 0 0 0
and their product (in either order) is the zero matrix.
Three.IV.2.31 Note that (S + T )(S − T ) = S 2 − ST + T S − T 2 , so a reasonable try is to look at
matrices that do not commute so that −ST ( and)
T S don’t ( cancel: ) with
1 2 5 6
S = T =
3 4 7 8
we have the desired inequality.
( )
( )
−56 −56
(S + T )(S − T ) =
S 2 − T 2 −60 −68
=
−88 −88
−76 −84
Three.IV.2.32 Because the identity map acts on the basis B as β ⃗ 1 ↦→ β ⃗ 1 , . . . , β ⃗ n ↦→ β ⃗ n , the representation
is this.
⎛
⎞
1 0 0 0
0 1 0 0
0 0 1 0
⎜
⎝
. ..
⎟
⎠
0 0 0 1
The second part of the question is obvious from Theorem 2.6.
Three.IV.2.33
Here are four solutions.
T =
⎛
⎞
( )
±1 0
0 ±1
Three.IV.2.34 (a) The vector space M 2×2 has dimension four. The set {T 4 , . . . , T, I} has five elements
and thus is linearly dependent.
(b) Where T is n × n, generalizing the argument from the prior item shows that there is such a
polynomial of degree n 2 or less, since {T n2 , . . . , T, I} is a n 2 +1-member subset of the n 2 -dimensional
space M n×n .
(c) First compute the powers
( √ ) ( ) ( √ )
T 2 1/2 − 3/2
= √ T 3 0 −1
=
T 4 = √ −1/2 − 3/2
3/2 1/2 1 0
3/2 −1/2
(observe that rotating by π/6 three times results in a rotation by π/2, which is indeed what T 3
represents). Then set c 4 T 4 + c 3 T 3 + c 2 T 2 + c 1 T + c 0 I equal to the zero matrix
( √ ) ( ) ( √ ) (√ ) ( )
√ −1/2 − 3/2 0 −1 1/2 − 3/2
c 4 + c 3/2 −1/2 1 0 3 + √ 3/2 −1/2
c 2 + √ 1 0
c 1 + c 3/2 1/2 1/2 3/2 0 1 0
)
=
⎛
⎞
⎛
⎞
(
0 0
0 0