Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 107
(the matrix has n columns because V is n-dimensional and it has only one row because R is onedimensional).
Then taking ⃗x to be the column vector that is the transpose of this matrix
⎛ ⎞
g 1
⎜
⃗x = . ⎟
⎝ . ⎠
g n
has the desired action.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
v 1 g 1 v 1
⎜
⃗v = . ⎟ ⎜
⎝ . ⎠ ↦→ . ⎟ ⎜ . ⎟
⎝ . ⎠ ⎝ . ⎠ = g 1 v 1 + · · · + g n v n
v n g n v n
(c) No. If ⃗x has any nonzero entries then h ⃗x cannot be the zero map (and if ⃗x is the zero vector then
h ⃗x can only be the zero map).
Three.III.2.22
See the following section.
Subsection Three.IV.1: Sums and Scalar Products
Three.IV.1.7
(a)
(e) Not defined.
( 7 0
) 6
9 1 6
(b)
( 12 −6
) −6
6 12 18
(c)
( ) 4 2
0 6
(d)
( ) −1 28
2 1
Three.IV.1.8 Represent the domain vector ⃗v ∈ V and the maps g, h: V → W with respect to bases
B, D in the usual way.
(a) The representation of (g + h) (⃗v) = g(⃗v) + h(⃗v)
(
(g1,1 v 1 + · · · + g 1,n v n ) ⃗ δ 1 + · · · + (g m,1 v 1 + · · · + g m,n v n ) ⃗ )
δ m
regroups
+ ( (h 1,1 v 1 + · · · + h 1,n v n ) ⃗ δ 1 + · · · + (h m,1 v 1 + · · · + h m,n v n ) ⃗ δ m
)
= ((g 1,1 + h 1,1 )v 1 + · · · + (g 1,1 + h 1,n )v n ) · ⃗δ 1 + · · · + ((g m,1 + h m,1 )v 1 + · · · + (g m,n + h m,n )v n ) · ⃗δ m
to the entry-by-entry sum of the representation of g(⃗v) and the representation of h(⃗v).
(b) The representation of (r · h) (⃗v) = r · (h(⃗v) )
r · ((h
1,1 v 1 + h 1,2 v 2 + · · · + h 1,n v n ) ⃗ δ 1 + · · · + (h m,1 v 1 + h m,2 v 2 + · · · + h m,n v n ) ⃗ δ m
)
= (rh 1,1 v 1 + · · · + rh 1,n v n ) · ⃗δ 1 + · · · + (rh m,1 v 1 + · · · + rh m,n v n ) · ⃗δ m
is the entry-by-entry multiple of r and the representation of h.
Three.IV.1.9 First, each of these properties is easy to check in an entry-by-entry way. For example,
writing
⎛
⎞ ⎛
⎞
g 1,1 . . . g 1,n
h 1,1 . . . h 1,n
⎜
⎟ ⎜
⎟
G = ⎝ . . ⎠ H = ⎝ . . ⎠
g m,1 . . . g m,n h m,1 . . . h m,n
then, by definition we have
⎛
⎞
⎛
⎞
g 1,1 + h 1,1 . . . g 1,n + h 1,n
h 1,1 + g 1,1 . . . h 1,n + g 1,n
⎜
G + H =
.
. ⎟
⎜
⎝ .
. ⎠ H + G =
.
. ⎟
⎝ .
. ⎠
g m,1 + h m,1 . . . g m,n + h m,n h m,1 + g m,1 . . . h m,n + g m,n
and the two are equal since their entries are equal g i,j + h i,j = h i,j + g i,j . That is, each of these is easy
to check by using Definition 1.3 alone.
However, each property is also easy to understand in terms of the represented maps, by applying
Theorem 1.5 as well as the definition.
(a) The two maps g + h and h + g are equal because g(⃗v) + h(⃗v) = h(⃗v) + g(⃗v), as addition is commutative
in any vector space. Because the maps are the same, they must have the same representative.
(b) As with the prior answer, except that here we apply that vector space addition is associative.
(c) As before, except that here we note that g(⃗v) + z(⃗v) = g(⃗v) + ⃗0 = g(⃗v).
(d) Apply that 0 · g(⃗v) = ⃗0 = z(⃗v).