Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
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100 <strong>Linear</strong> <strong>Algebra</strong>, by Hefferon<br />
(c) We first find the image of each member of B, and then represent those images with respect to<br />
D. For the first step,<br />
(<br />
we<br />
)<br />
can<br />
(<br />
use the<br />
)<br />
matrix<br />
(<br />
developed<br />
) (<br />
earlier.<br />
) ( ( )<br />
1 −1 3 1 −4<br />
1 −4<br />
Rep E2<br />
( ) =<br />
=<br />
so t( ) =<br />
−1 0 0 −1 0<br />
−1)<br />
0<br />
E 2,E 2 E 2 E 2<br />
Actually, for the second member of B there is no need to apply the matrix because the problem<br />
statement gives its image.<br />
( (<br />
1 2<br />
t( ) =<br />
1)<br />
0)<br />
Now representing those images<br />
(<br />
with<br />
)<br />
respect<br />
( )<br />
to D is routine.<br />
( ( )<br />
−4 −1<br />
2 1/2<br />
Rep D ( ) = and Rep<br />
0 2<br />
D ( ) =<br />
0)<br />
−1<br />
D<br />
D<br />
Thus, the matrix is this.<br />
( ) −1 1/2<br />
Rep B,D (t) =<br />
2 −1<br />
B,D<br />
(d) We know the images of the members of the domain’s basis from the prior item.<br />
( ( ) ( ( 1 −4 1 2<br />
t( ) =<br />
t( ) =<br />
−1)<br />
0 1)<br />
0)<br />
We can compute the representation<br />
( )<br />
of those<br />
( )<br />
images with respect<br />
(<br />
to the<br />
(<br />
codomain’s basis.<br />
−4 −2<br />
2 1<br />
Rep B ( ) = and Rep<br />
0 −2<br />
B ( ) =<br />
0)<br />
1)<br />
B<br />
B<br />
Thus this is the matrix.<br />
( )<br />
−2 1<br />
Rep B,B (t) =<br />
−2 1<br />
B,B<br />
Three.III.1.23 (a) The images of the members of the domain’s basis are<br />
⃗β 1 ↦→ h( β ⃗ 1 ) β2 ⃗ ↦→ h( β ⃗ 2 ) . . . βn ⃗ ↦→ h( β ⃗ n )<br />
and those images are represented with respect to the codomain’s basis in this way.<br />
⎛ ⎞<br />
⎛ ⎞<br />
⎛ ⎞<br />
1<br />
0<br />
0<br />
Rep h(B) ( h( β ⃗ 0<br />
1 ) ) = ⎜<br />
⎝<br />
⎟<br />
. ⎠ Rep h(B)( h( β ⃗ 1<br />
2 ) ) = ⎜<br />
⎝<br />
⎟<br />
. ⎠ . . . Rep h(B)( h( β ⃗ 0<br />
n ) ) = ⎜<br />
⎝<br />
⎟<br />
. ⎠<br />
0<br />
0<br />
1<br />
Hence, the matrix is the identity.<br />
⎛<br />
⎞<br />
1 0 . . . 0<br />
0 1 0<br />
Rep B,h(B) (h) = ⎜<br />
⎝<br />
. ..<br />
⎟<br />
⎠<br />
0 0 1<br />
(b) Using the matrix in the prior item, the representation is this.<br />
⎛ ⎞<br />
1<br />
⎜<br />
Rep h(B) ( h(⃗v) ) = ⎝c<br />
.<br />
⎟<br />
⎠<br />
c n<br />
h(B)<br />
Three.III.1.24 The product<br />
⎛ ⎞<br />
⎛<br />
⎞ 0 ⎛ ⎞<br />
h 1,1 . . . h 1,i . . . h 1,n<br />
.<br />
h 1,i<br />
h 2,1 . . . h 2,i . . . h 2,n<br />
.<br />
⎜<br />
⎟<br />
⎝ .<br />
⎠<br />
1<br />
h 2,i<br />
= ⎜ ⎟<br />
⎜ ⎟ ⎝ . ⎠<br />
⎝<br />
h m,1 . . . h m,i . . . h 1,n<br />
. ⎠<br />
h m,i<br />
0<br />
gives the i-th column of the matrix.<br />
Three.III.1.25 (a) The images of the basis vectors for the domain are cos x d/dx<br />
↦−→ − sin x and sin x d/dx<br />
↦−→<br />
cos x. Representing those with respect to the codomain’s basis (again, B) and adjoining the representations<br />
gives this matrix.<br />
Rep B,B ( d ( )<br />
0 1<br />
dx ) = −1 0<br />
B,B
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