Linear Algebra Exercises-n-Answers.pdf

180 Linear Algebra, by Hefferon
Five.II.1.8
One possible choice of the bases is
( ( ) 1 −1
B = 〈 , 〉 D = E
2)
1
2 = 〈
( 1
0)
( 0
, 〉
1)
(this B is suggested by the map description). To find the matrix T = Rep B,B (t), solve the relations
( ) ( ) ( ( ) ( ) ( )
1 −1 3 1 −1 −1
c 1 + c
2 2 = ĉ
1 0)
1 + ĉ
2 2 =
1 2
to get c 1 = 1, c 2 = −2, ĉ 1 = 1/3 and ĉ 2 = 4/3.
Rep B,B (t) =
( 1
) 1/3
−2 4/3
Finding Rep D,D (t) involves a bit more computation. We first find t(⃗e 1 ). The relation
( ) ( ) (
1 −1 1
c 1 + c
2 2 =
1 0)
gives c 1 = 1/3 and c 2 = −2/3, and so
making
Rep B (⃗e 1 ) =
( ) 1/3
−2/3
B
( ) ( ) ( )
1 1/3 1/3 1/9
Rep B (t(⃗e 1 )) =
=
−2 4/3 −2/3 −14/9
B,B
B
B
and hence t acts on the first basis vector ⃗e 1 in this way.
( ( ) ( )
1 −1 5/3
t(⃗e 1 ) = (1/9) · − (14/9) · =
2)
1 −4/3
The computation for t(⃗e 2 ) is similar. The relation
( ) ( ) (
1 −1 0
c 1 + c
2 2 =
1 1)
gives c 1 = 1/3 and c 2 = 1/3, so
making
Rep B (⃗e 1 ) =
( ) 1/3
1/3
B
( )
1 1/3
Rep B (t(⃗e 1 )) =
−2 4/3
B,B
and hence t acts on the second basis vector ⃗e 2 in this way.
t(⃗e 2 ) = (4/9) ·
(
1
2)
− (2/9) ·
( ) ( )
1/3 4/9
=
1/3 −2/9
B
B
(
−1
1
)
=
( )
2/3
2/3
Therefore
( )
5/3 2/3
Rep D,D (t) =
−4/3 2/3
and these are the change of basis matrices.
( )
1 −1
P = Rep B,D (id) =
P −1 = ( Rep
2 1
B,D (id) ) ( ) −1
−1 1 −1 = =
2 1
The check of these computations is routine.
( ) ( ) ( ) ( )
1 −1 1 1/3 1/3 1/3 5/3 2/3
=
2 1 −2 4/3 −2/3 1/3 −4/3 2/3
(
1/3
)
1/3
−2/3 1/3
Five.II.1.9 The only representation of a zero map is a zero matrix, no matter what the pair of bases
Rep B,D (z) = Z, and so in particular for any single basis B we have Rep B,B (z) = Z. The case of the
identity is related, but slightly different: the only representation of the identity map, with respect to
any B, B, is the identity Rep B,B (id) = I. (Remark: of course, we have seen examples where B ≠ D
and Rep B,D (id) ≠ I — in fact, we have seen that any nonsingular matrix is a representation of the
identity map with respect to some B, D.)
Five.II.1.10 No. If A = P BP −1 then A 2 = (P BP −1 )(P BP −1 ) = P B 2 P −1 .
Five.II.1.11 Matrix similarity is a special case of matrix equivalence (if matrices are similar then they
are matrix equivalent) and matrix equivalence preserves nonsingularity.