Linear Algebra Exercises-n-Answers.pdf

122 Linear Algebra, by Hefferon
we can either solve the
(
system
( ) ( ) ( ) ( ) ( )
1 x1 x2 0 x1 x2
= 5 + 0
= 0 + 4
0)
y 1 y 1 1 y 1 y 1
or else just spot the answer (thinking of the
(
proof
)
of
(
Lemma
)
1.4).
1/5 0
D = 〈 , 〉
0 1/4
(b) Yes, this matrix is nonsingular and so changes
( )
bases.
( )
To calculate D, we proceed as above with
x1 x2
D = 〈 , 〉
y 1 y 2
to solve (
1
0)
( ) ( )
x1 x2
= 2 + 3
y 1 y 1
and
(
0
1)
( ) ( )
x1 x2
= 1 + 1
y 1 y 1
and get this.
( ) (
−1 1
D = 〈 , 〉
3 −2)
(c) No, this matrix does not change bases because it is nonsingular.
(d) Yes, this matrix changes bases because it is nonsingular. The calculation of the changed-to basis
is as above.
( ) ( )
1/2 1/2
D = 〈 , 〉
−1/2 1/2
Three.V.1.11 This question has many different solutions. One way to proceed is to make up any basis
B for any space, and then compute the appropriate D (necessarily for the same space, of course).
Another, easier, way to proceed is to fix the codomain as R 3 and the codomain basis as E 3 . This way
(recall that the representation of any vector with respect to the standard basis is just the vector itself),
we have this.
⎛
B = 〈 ⎝ 3 ⎞ ⎛
2⎠ , ⎝ 1
⎞ ⎛
−1⎠ , ⎝ 4 ⎞
1⎠〉 D = E 3
0 0 4
Three.V.1.12 Checking that B = 〈2 sin(x) + cos(x), 3 cos(x)〉 is a basis is routine. Call the natural
basis D. To compute the change of basis matrix Rep B,D (id) we must find Rep D (2 sin(x) + cos(x)) and
Rep D (3 cos(x)), that is, we need x 1 , y 1 , x 2 , y 2 such that these equations hold.
x 1 · sin(x) + y 1 · cos(x) = 2 sin(x) + cos(x)
x 2 · sin(x) + y 2 · cos(x) = 3 cos(x)
Obviously this is the answer.
( )
2 0
Rep B,D (id) =
1 3
For the change of basis matrix in the other direction we could look for Rep B (sin(x)) and Rep B (cos(x))
by solving these.
w 1 · (2 sin(x) + cos(x)) + z 1 · (3 cos(x)) = sin(x)
w 2 · (2 sin(x) + cos(x)) + z 2 · (3 cos(x)) = cos(x)
An easier method is to find the inverse
(
of the
)
matrix found above.
−1
2 0
Rep D,B (id) =
= 1 ( ) ( )
3 0 1/2 0
1 3 6 · =
−1 2 −1/6 1/3
Three.V.1.13 We start by taking the inverse of the matrix, that is, by deciding what is the inverse to
the map of interest.
( ) ( )
Rep D,E2 (id)Rep D,E2 (id) −1 1
− cos(2θ) − sin(2θ) cos(2θ) sin(2θ)
=
− cos 2 (2θ) − sin 2 (2θ) ·
=
− sin(2θ) cos(2θ) sin(2θ) − cos(2θ)
This is more tractable than the representation the other way because this matrix is the concatenation
of these two column vectors ( )
( )
Rep E2
( ⃗ cos(2θ)
δ 1 ) =
Rep
sin(2θ)
E2
( ⃗ sin(2θ)
δ 2 ) =
− cos(2θ)
and representations with respect to E
( 2 are transparent.
) ( )
⃗ cos(2θ)
δ1 = ⃗ sin(2θ)
δ2 =
sin(2θ)
− cos(2θ)
This pictures the action of the map that transforms D to E 2 (it is, again, the inverse of the map that
is the answer to this question). The line lies at an angle θ to the x axis.