Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 95
for x 1 , . . . , x n ∈ R. This function h from V to R
c 1β1 ⃗ + · · · + c nβn
⃗
h
↦−→ c 1 x 1 + · · · + c n x n
is easily seen to be linear, and to be mapped by Φ to the given vector in R n , so Φ is onto.
The map Φ also preserves structure: where
c 1β1 ⃗ + · · · + c nβn ⃗
h
↦−→
1
c1 h 1 ( β ⃗ 1 ) + · · · + c n h 1 ( β ⃗ n )
c 1β1 ⃗ + · · · + c nβn ⃗
h
↦−→
2
c1 h 2 ( β ⃗ 1 ) + · · · + c n h 2 ( β ⃗ n )
we have
(r 1 h 1 + r 2 h 2 )(c 1β1 ⃗ + · · · + c nβn ⃗ ) = c 1 (r 1 h 1 ( β ⃗ 1 ) + r 2 h 2 ( β ⃗ 1 )) + · · · + c n (r 1 h 1 ( β ⃗ n ) + r 2 h 2 ( β ⃗ n ))
= r 1 (c 1 h 1 ( ⃗ β 1 ) + · · · + c n h 1 ( ⃗ β n )) + r 2 (c 1 h 2 ( ⃗ β 1 ) + · · · + c n h 2 ( ⃗ β n ))
so Φ(r 1 h 1 + r 2 h 2 ) = r 1 Φ(h 1 ) + r 2 Φ(h 2 ).
Three.II.2.43
V to W
Let h: V → W be linear and fix a basis 〈 ⃗ β 1 , . . . , ⃗ β n 〉 for V . Consider these n maps from
h 1 (⃗v) = c 1 · h( ⃗ β 1 ), h 2 (⃗v) = c 2 · h( ⃗ β 2 ), . . . , h n (⃗v) = c n · h( ⃗ β n )
for any ⃗v = c 1
⃗ β1 + · · · + c n
⃗ βn . Clearly h is the sum of the h i ’s. We need only check that each h i is
linear: where ⃗u = d 1
⃗ β1 + · · · + d n
⃗ βn we have h i (r⃗v + s⃗u) = rc i + sd i = rh i (⃗v) + sh i (⃗u).
Three.II.2.44 Either yes (trivially) or no (nearly trivially).
If V ‘is homomorphic to’ W is taken to mean there is a homomorphism from V into (but not
necessarily onto) W , then every space is homomorphic to every other space as a zero map always
exists.
If V ‘is homomorphic to’ W is taken to mean there is an onto homomorphism from V to W then the
relation is not an equivalence. For instance, there is an onto homomorphism from R 3 to R 2 (projection
is one) but no homomorphism from R 2 onto R 3 by Corollary 2.17, so the relation is not reflexive. ∗
Three.II.2.45 That they form the chains is obvious. For the rest, we show here that R(t j+1 ) = R(t j )
implies that R(t j+2 ) = R(t j+1 ). Induction then applies.
Assume that R(t j+1 ) = R(t j ). Then t: R(t j+1 ) → R(t j+2 ) is the same map, with the same
domain, as t: R(t j ) → R(t j+1 ). Thus it has the same range: R(t j+2 ) = R(t j+1 ).
Subsection Three.III.1: Representing Linear Maps with Matrices
⎛
Three.III.1.11 (a) ⎝ 1 · 2 + 3 · 1 + 1 · 0 ⎞ ⎛
0 · 2 + (−1) · 1 + 2 · 0⎠ = ⎝ 5 ⎞
⎛
−1⎠ (b) Not defined. (c) ⎝ 0 ⎞
0⎠
1 · 2 + 1 · 1 + 0 · 0 3
0
( ) ( ) ( 2 · 4 + 1 · 2 10
4
Three.III.1.12 (a)
= (b) (c) Not defined.
3 · 4 − (1/2) · 2 11
1)
Three.III.1.13 Matrix-vector multiplication gives rise to a linear system.
2x + y + z = 8
y + 3z = 4
x − y + 2z = 4
Gaussian reduction shows that z = 1, y = 1, and x = 3.
Three.III.1.14 Here are two ways to get the answer.
First, obviously 1 − 3x + 2x 2 = 1 · 1 − 3 · x + 2 · x 2 , and so we can apply the general property of
preservation of combinations to get h(1−3x+2x 2 ) = h(1·1−3·x+2·x 2 ) = 1·h(1)−3·h(x)+2·h(x 2 ) =
1 · (1 + x) − 3 · (1 + 2x) + 2 · (x − x 3 ) = −2 − 3x − 2x 3 .
The other way uses the computation scheme developed in this subsection. Because we know where
these elements of the space go, we consider this basis B = 〈1, x, x 2 〉 for the domain. Arbitrarily, we
can take D = 〈1, x, x 2 , x 3 〉 as a basis for the codomain. With those choices, we have that
⎛ ⎞
1 1 0
Rep B,D (h) = ⎜1 2 1
⎟
⎝0 0 0 ⎠
0 0 −1
∗ More information on equivalence relations is in the appendix.
B,D