Linear Algebra Exercises-n-Answers.pdf

94 Linear Algebra, by Hefferon
Three.II.2.39
This is a simple calculation.
h([S]) = {h(c 1 ⃗s 1 + · · · + c n ⃗s n ) ∣ ∣ c 1 , . . . , c n ∈ R and ⃗s 1 , . . . , ⃗s n ∈ S}
= {c 1 h(⃗s 1 ) + · · · + c n h(⃗s n ) ∣ ∣ c 1 , . . . , c n ∈ R and ⃗s 1 , . . . , ⃗s n ∈ S}
= [h(S)]
Three.II.2.40 (a) We will show that the two sets are equal h −1 ( ⃗w) = {⃗v + ⃗n ∣ ⃗n ∈ N (h)} by mutual
inclusion. For the {⃗v + ⃗n ∣ ⃗n ∈ N (h)} ⊆ h −1 ( ⃗w) direction, just note that h(⃗v + ⃗n) = h(⃗v) +
h(⃗n) equals ⃗w, and so any member of the first set is a member of the second. For the h −1 ( ⃗w) ⊆
{⃗v + ⃗n ∣ ⃗n ∈ N (h)} direction, consider ⃗u ∈ h −1 ( ⃗w). Because h is linear, h(⃗u) = h(⃗v) implies that
h(⃗u − ⃗v) = ⃗0. We can write ⃗u − ⃗v as ⃗n, and then we have that ⃗u ∈ {⃗v + ⃗n ∣ ⃗n ∈ N (h)}, as desired,
because ⃗u = ⃗v + (⃗u − ⃗v).
(b) This check is routine.
(c) This is immediate.
(d) For the linearity check, briefly, where c, d are scalars and ⃗x, ⃗y ∈ R n have components x 1 , . . . , x n
and y 1 , . . . , y n , we have this.
⎛
⎞
a 1,1 (cx 1 + dy 1 ) + · · · + a 1,n (cx n + dy n )
⎜
⎟
h(c · ⃗x + d · ⃗y) = ⎝
.
⎠
a m,1 (cx 1 + dy 1 ) + · · · + a m,n (cx n + dy n )
⎛
⎛
=
⎜
⎝
⎞
⎞
a 1,1 cx 1 + · · · + a 1,n cx n a 1,1 dy 1 + · · · + a 1,n dy n
.
⎟ ⎜
.
⎠ +
.
⎟
⎝
.
⎠
a m,1 cx 1 + · · · + a m,n cx n a m,1 dy 1 + · · · + a m,n dy n
= c · h(⃗x) + d · h(⃗y)
The appropriate conclusion is that General = Particular + Homogeneous.
(e) Each power of the derivative is linear because of the rules
d k
dk dk
(f(x) + g(x)) = f(x) +
dxk dxk dx k g(x) and d k
dk
rf(x) = r
dxk dx k f(x)
from calculus. Thus the given map is a linear transformation of the space because any linear
combination of linear maps is also a linear map by Lemma 1.16. The appropriate conclusion is
General = Particular + Homogeneous, where the associated homogeneous differential equation has
a constant of 0.
Three.II.2.41 Because the rank of t is one, the rangespace of t is a one-dimensional set. Taking 〈h(⃗v)〉
as a basis (for some appropriate ⃗v), we have that for every ⃗w ∈ V , the image h( ⃗w) ∈ V is a multiple
of this basis vector — associated with each ⃗w there is a scalar c ⃗w such that t( ⃗w) = c ⃗w t(⃗v). Apply t to
both sides of that equation and take r to be c t(⃗v)
t ◦ t( ⃗w) = t(c ⃗w · t(⃗v)) = c ⃗w · t ◦ t(⃗v) = c ⃗w · c t(⃗v) · t(⃗v) = c ⃗w · r · t(⃗v) = r · c ⃗w · t(⃗v) = r · t( ⃗w)
to get the desired conclusion.
Three.II.2.42 Fix a basis 〈 β ⃗ 1 , . . . , β ⃗ n 〉 for V . We shall prove that this map
⎛
h( ⃗ ⎞
β 1 )
h ↦−→
Φ ⎜ . ⎟
⎝ . ⎠
h( β ⃗ n )
is an isomorphism from V ∗ to R n .
To see that Φ is one-to-one, assume that h 1 and h 2 are members of V ∗ such that Φ(h 1 ) = Φ(h 2 ).
Then
⎛ ⎞ ⎛ ⎞
⎜
⎝
h 1 ( ⃗ β 1 )
.
h 1 ( ⃗ β n )
⎟
⎠ =
⎜
⎝
h 2 ( ⃗ β 1 )
.
h 2 ( ⃗ β n )
and consequently, h 1 ( β ⃗ 1 ) = h 2 ( β ⃗ 1 ), etc. But a homomorphism is determined by its action on a basis,
so h 1 = h 2 , and therefore Φ is one-to-one.
To see that Φ is onto, consider
⎛
⎜
⎝
⎞
x 1
. ⎟
. ⎠
x n
⎟
⎠