Linear Algebra Exercises-n-Answers.pdf

88 Linear Algebra, by Hefferon
Three.II.1.29 Verifying that it is linear is routine.
⎛
h(c 1 · ⎝ x ⎞ ⎛
1
y 1
⎠ + c 2 · ⎝ x ⎞ ⎛
2
y 2
⎠) = h( ⎝ c ⎞
1x 1 + c 2 x 2
c 1 y 1 + c 2 y 2
⎠)
z 1 z 2 c 1 z 1 + c 2 z 2
= 3(c 1 x 1 + c 2 x 2 ) − (c 1 y 1 + c 2 y 2 ) − (c 1 z 1 + c 2 z 2 )
= c 1 · (3x 1 − y 1 − z 1 ) + c 2 · (3x 2 − y 2 − z 2 )
⎛
= c 1 · h( ⎝ x ⎞ ⎛
1
y 1
⎠) + c 2 · h( ⎝ x ⎞
2
y 2
⎠)
z 1 z 2
The natural guess at a generalization is that for any fixed ⃗ k ∈ R 3 the map ⃗v ↦→ ⃗v ⃗ k is linear. This
statement is true. It follows from properties of the dot product we have seen earlier: (⃗v+⃗u) ⃗ k = ⃗v ⃗ k+⃗u ⃗ k
and (r⃗v) ⃗ k = r(⃗v ⃗ k). (The natural guess at a generalization of this generalization, that the map from
R n to R whose action consists of taking the dot product of its argument with a fixed vector ⃗ k ∈ R n is
linear, is also true.)
Three.II.1.30 Let h: R 1 → R 1 be linear. A linear map is determined by its action on a basis, so fix the
basis 〈1〉 for R 1 . For any r ∈ R 1 we have that h(r) = h(r · 1) = r · h(1) and so h acts on any argument
r by multiplying it by the constant h(1). If h(1) is not zero then the map is a correspondence — its
inverse is division by h(1) — so any nontrivial transformation of R 1 is an isomorphism.
This projection map is an example that shows that not every transformation of R n acts via multiplication
by a constant when n > 1, including when n = 2.
⎛ ⎞ ⎛ ⎞
x 1 x 1
x 2
⎜
⎝
⎟
. ⎠ ↦→ 0
⎜
⎝
⎟
. ⎠
x n 0
Three.II.1.31 (a) Where c and d are scalars, we have this.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1 y 1 cx 1 + dy 1
⎜
h(c · ⎝
⎟ ⎜
. ⎠ + d · ⎝
⎟ ⎜
. ⎠) = h( ⎝
⎟
. ⎠)
x n y n cx n + dy n
⎛
⎞
a 1,1 (cx 1 + dy 1 ) + · · · + a 1,n (cx n + dy n )
⎜
= ⎝
⎟
.
⎠
a m,1 (cx 1 + dy 1 ) + · · · + a m,n (cx n + dy n )
⎛
⎞ ⎛
⎞
a 1,1 x 1 + · · · + a 1,n x n a 1,1 y 1 + · · · + a 1,n y n
⎜
⎟ ⎜
⎟
= c · ⎝ . ⎠ + d · ⎝ . ⎠
a m,1 x 1 + · · · + a m,n x n a m,1 y 1 + · · · + a m,n y n
⎛ ⎞ ⎛ ⎞
x 1
y 1
⎜
= c · h( . ⎟ ⎜
⎝ . ⎠) + d · h( . ⎟
⎝ . ⎠)
x n y n
(b) Each power i of the derivative operator is linear because of these rules familiar from calculus.
d i
di di
( f(x) + g(x) ) = f(x) +
dxi dxi dx i g(x) and d i
dx i r · f(x) = r · d i
dx i f(x)
Thus the given map is a linear transformation of P n because any linear combination of linear maps
is also a linear map.
Three.II.1.32 (This argument has already appeared, as part of the proof that isomorphism is an equivalence.)
Let f : U → V and g : V → W be linear. For any ⃗u 1 , ⃗u 2 ∈ U and scalars c 1 , c 2 combinations
are preserved.
g ◦ f(c 1 ⃗u 1 + c 2 ⃗u 2 ) = g( f(c 1 ⃗u 1 + c 2 ⃗u 2 ) ) = g( c 1 f(⃗u 1 ) + c 2 f(⃗u 2 ) )
= c 1 · g(f(⃗u 1 )) + c 2 · g(f(⃗u 2 )) = c 1 · g ◦ f(⃗u 1 ) + c 2 · g ◦ f(⃗u 2 )
Three.II.1.33 (a) Yes. The set of ⃗w ’s cannot be linearly independent if the set of ⃗v ’s is linearly
dependent because any nontrivial relationship in the domain ⃗0 V = c 1 ⃗v 1 + · · · + c n ⃗v n would give a