Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 89
nontrivial relationship in the range h(⃗0 V ) = ⃗0 W = h(c 1 ⃗v 1 + · · · + c n ⃗v n ) = c 1 h(⃗v 1 ) + · · · + c n h(⃗v n ) =
c 1 ⃗w + · · · + c n ⃗w n .
(b) Not necessarily. For instance, the transformation of R 2 given by
( ( )
x x + y
↦→
y)
x + y
sends this linearly independent set in the domain to a linearly dependent image.
( ( ( (
1 1 1 2
{⃗v 1 , ⃗v 2 } = { , } ↦→ { , } = { ⃗w
0)
1)
1)
2)
1 , ⃗w 2 }
(c) Not necessarily. An example is the projection map π : R 3 → R 2
⎛
⎝ x ⎞
(
y⎠ x
↦→
y)
z
and this set that does not span the domain but maps to a set that does span the codomain.
⎛
{ ⎝ 1 ⎞ ⎛
0⎠ , ⎝ 0 ⎞
( (
1⎠} ↦−→ π 1 0
{ , }
0)
1)
0 0
(d) Not necessarily. For instance, the injection map ι: R 2 → R 3 sends the standard basis E 2 for the
domain to a set that does not span the codomain. (Remark. However, the set of ⃗w’s does span the
range. A proof is easy.)
Three.II.1.34 Recall that the entry in row i and column j of the transpose of M is the entry m j,i
from row j and column i of M. Now, the check is routine.
⎛
⎞ ⎛
⎞ trans ⎛
⎞trans
.
[r · ⎜
⎝· · · a i,j · · · ⎟
⎠ + s · .
⎜
⎝· · · b i,j · · · ⎟
⎠ ] .
= ⎜
⎝· · · ra i,j + sb i,j · · · ⎟
⎠
.
.
.
.
.
.
⎛
⎞
.
= ⎜
⎝· · · ra j,i + sb j,i · · · ⎟
⎠
.
⎛
⎞ ⎛
⎞
.
.
.
= r · ⎜
⎝
· · · a j,i · · · ⎟
⎠ + s · .
⎜
⎝
· · · b j,i · · · ⎟
⎠
The domain is M m×n while the codomain is M n×m .
Three.II.1.35
.
⎛
⎞
.
= r · ⎜
⎝· · · a j,i · · · ⎟
⎠
.
(a) For any homomorphism h: R n → R m we have
trans
.
⎛
⎞
.
+ s · ⎜
⎝· · · b j,i · · · ⎟
⎠
.
h(l) = {h(t · ⃗u + (1 − t) · ⃗v) ∣ ∣ t ∈ [0..1]} = {t · h(⃗u) + (1 − t) · h(⃗v) ∣ ∣ t ∈ [0..1]}
which is the line segment from h(⃗u) to h(⃗v).
(b) We must show that if a subset of the domain is convex then its image, as a subset of the range, is
also convex. Suppose that C ⊆ R n is convex and consider its image h(C). To show h(C) is convex
we must show that for any two of its members, ⃗ d 1 and ⃗ d 2 , the line segment connecting them
l = {t · ⃗d 1 + (1 − t) · ⃗d 2
∣ ∣ t ∈ [0..1]}
is a subset of h(C).
Fix any member ˆt · ⃗d 1 + (1 − ˆt) · ⃗d 2 of that line segment. Because the endpoints of l are in the
image of C, there are members of C that map to them, say h(⃗c 1 ) = ⃗ d 1 and h(⃗c 2 ) = ⃗ d 2 . Now, where
ˆt is the scalar that is fixed in the first sentence of this paragraph, observe that h(ˆt ·⃗c 1 + (1 − ˆt) ·⃗c 2 ) =
ˆt · h(⃗c 1 ) + (1 − ˆt) · h(⃗c 2 ) = ˆt · ⃗d 1 + (1 − ˆt) · ⃗d 2 Thus, any member of l is a member of h(C), and so
h(C) is convex.
trans