Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 81
gives that cos φ = 1/ √ 1 + k 2 and sin φ = k/ √ 1 + k 2 . Now,
cos(2φ − θ) = cos(2φ) cos θ + sin(2φ) sin θ
= ( cos 2 φ − sin 2 φ ) cos θ + (2 sin φ cos φ) sin θ
(
) (
)
1
k
k 1
= ( √
1 + k
2 )2 − ( √
1 + k
2 )2 cos θ + 2√ √ sin θ
1 + k
2 1 + k
2
( ) ( )
1 − k
2
2k
=
1 + k 2 cos θ +
1 + k 2 sin θ
and thus the first component of the image vector is this.
r · cos(2φ − θ) = 1 − k2
1 + k 2 · x + 2k
1 + k 2 · y
A similar calculation shows that the second component of the image vector is this.
r · sin(2φ − θ) =
2k
1 + k 2 · x − 1 − k2
1 + k 2 · y
With this algebraic description of the action of f
( ( l
)
x f l (1 − k ↦−→
y)
2 /1 + k 2 ) · x + (2k/1 + k 2 ) · y
(2k/1 + k 2 ) · x − (1 − k 2 /1 + k 2 ) · y
checking that it preserves structure is routine.
Three.I.1.30 First, the map p(x) ↦→ p(x + k) doesn’t count because it is a version of p(x) ↦→ p(x − k).
Here is a correct answer (many others are also correct): a 0 +a 1 x+a 2 x 2 ↦→ a 2 +a 0 x+a 1 x 2 . Verification
that this is an isomorphism is straightforward.
Three.I.1.31 (a) For the ‘only if’ half, let f : R 1 → R 1 to be an isomorphism. Consider the basis
〈1〉 ⊆ R 1 . Designate f(1) by k. Then for any x we have that f(x) = f(x · 1) = x · f(1) = xk, and
so f’s action is multiplication by k. To finish this half, just note that k ≠ 0 or else f would not be
one-to-one.
For the ‘if’ half we only have to check that such a map is an isomorphism when k ≠ 0. To check
that it is one-to-one, assume that f(x 1 ) = f(x 2 ) so that kx 1 = kx 2 and divide by the nonzero factor
k to conclude that x 1 = x 2 . To check that it is onto, note that any y ∈ R 1 is the image of x = y/k
(again, k ≠ 0). Finally, to check that such a map preserves combinations of two members of the
domain, we have this.
f(c 1 x 1 + c 2 x 2 ) = k(c 1 x 1 + c 2 x 2 ) = c 1 kx 1 + c 2 kx 2 = c 1 f(x 1 ) + c 2 f(x 2 )
(b) By the prior item, f’s action is x ↦→ (7/3)x. Thus f(−2) = −14/3.
(c) For the ‘only if’ half, assume that f : R 2 → R 2 is an automorphism. Consider the standard basis
E 2 for R 2 . Let
( ( a b
f(⃗e 1 ) = and f(⃗e
c)
2 ) = .
d)
Then the action of f on any vector is determined by by its action on the two basis vectors.
( ( ( ( )
x a b ax + by
f( ) = f(x · ⃗e
y)
1 + y · ⃗e 2 ) = x · f(⃗e 1 ) + y · f(⃗e 2 ) = x · + y · =
c)
d)
cx + dy
To finish this half, note that if ad − bc = 0, that is, if f(⃗e 2 ) is a multiple of f(⃗e 1 ), then f is not
one-to-one.
For ‘if’ we must check that the map is an isomorphism, under the condition that ad − bc ≠ 0.
The structure-preservation check is easy; we will here show that f is a correspondence. For the
argument that the map is one-to-one, assume this.
f(
(
x1
y 1
)
) = f(
(
x2
y 2
)
) and so
Then, because ad − bc ≠ 0, the resulting system
a(x 1 − x 2 ) + b(y 1 − y 2 ) = 0
c(x 1 − x 2 ) + d(y 1 − y 2 ) = 0
( ) ( )
ax1 + by 1 ax2 + by
=
2
cx 1 + dy 1 cx 2 + dy 2
has a unique solution, namely the trivial one x 1 − x 2 = 0 and y 1 − y 2 = 0 (this follows from the
hint).
The argument that this map is onto is closely related — this system
ax 1 + by 1 = x
cx 1 + dy 1 = y