Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 109
( ) ( ) ( ) ( ) ( )
1 −2
1 −2 −2 3 6 1
−18 17
Three.IV.2.15 (a)
(b)
=
(c)
10 4
10 4 −4 1 −36 34 −24 16
( ) ( ) ( )
1 −1 −18 17 6 1
(d)
=
2 0 −24 16 −36 34
Three.IV.2.16 (a) Yes. (b) Yes. (c) No. (d) No.
Three.IV.2.17 (a) 2×1 (b) 1×1 (c) Not defined. (d) 2×2
Three.IV.2.18 We have
h 1,1 · (g 1,1 y 1 + g 1,2 y 2 ) + h 1,2 · (g 2,1 y 1 + g 2,2 y 2 ) + h 1,3 · (g 3,1 y 1 + g 3,2 y 2 ) = d 1
h 2,1 · (g 1,1 y 1 + g 1,2 y 2 ) + h 2,2 · (g 2,1 y 1 + g 2,2 y 2 ) + h 2,3 · (g 3,1 y 1 + g 3,2 y 2 ) = d 2
which, after expanding and regrouping about the y’s yields this.
(h 1,1 g 1,1 + h 1,2 g 2,1 + h 1,3 g 3,1 )y 1 + (h 1,1 g 1,2 + h 1,2 g 2,2 + h 1,3 g 3,2 )y 2 = d 1
(h 2,1 g 1,1 + h 2,2 g 2,1 + h 2,3 g 3,1 )y 1 + (h 2,1 g 1,2 + h 2,2 g 2,2 + h 2,3 g 3,2 )y 2 = d 2
The starting system, and the system used for the substitutions, can be expressed in matrix language.
( ) ⎛
h1,1 h 1,2 h 1,3 ⎝ x ⎞ ⎛
1
x
h 2,1 h 2,2 h 2
⎠ = H ⎝ x ⎞
⎛
( )
1
x 2
⎠ d1
= ⎝ g ⎞
⎛
1,1 g 1,2
( ) ( )
g
2,3 d 2,1 g 2,2
⎠ y1 y1
= G = ⎝ x ⎞
1
x
x 3 x 2 y
3 g 2 y 2
⎠
2
3,2 x 3
With this, the substitution is ⃗ d = H⃗x = H(G⃗y) = (HG)⃗y.
Three.IV.2.19 Technically, no. The dot product operation yields a scalar while the matrix product
yields a 1×1 matrix. However, we usually will ignore the distinction.
Three.IV.2.20 The action of d/dx on B is 1 ↦→ 0, x ↦→ 1, x 2 ↦→ 2x, . . . and so this is its (n+1)×(n+1)
matrix representation.
⎛
⎞
0 1 0 0
Rep B,B ( d
0 0 2 0
dx ) = . ..
⎜
⎟
⎝0 0 0 n⎠
0 0 0 0
The product of this matrix with itself is defined because the matrix is square.
⎛
⎞
⎛
⎞2
0 0 2 0 0
0 1 0 0
0 0 2 0
0 0 0 6 0
. ..
. ..
=
⎜
⎟
⎝0 0 0 n⎠
⎜0 0 0 n(n − 1)
⎟
⎝0 0 0 0 ⎠
0 0 0 0
0 0 0 0
The map so represented is the composition
which is the second derivative operation.
p
d
dx
↦−→ d p
dx
g 3,1
d
dx
↦−→ d2 p
dx 2
Three.IV.2.21 It is true for all one-dimensional spaces. Let f and g be transformations of a onedimensional
space. We must show that g ◦ f (⃗v) = f ◦ g (⃗v) for all vectors. Fix a basis B for the space
and then the transformations are represented by 1×1 matrices.
F = Rep B,B (f) = ( )
f 1,1 G = Rep B,B (g) = ( )
g 1,1
Therefore, the compositions can be represented as GF and F G.
GF = Rep B,B (g ◦ f) = ( g 1,1 f 1,1
)
F G = Rep B,B (f ◦ g) = ( f 1,1 g 1,1
)
These two matrices are equal and so the compositions have the same effect on each vector in the space.
Three.IV.2.22
It would not represent linear map composition; Theorem 2.6 would fail.
Three.IV.2.23 Each follows easily from the associated map fact. For instance, p applications of the
transformation h, following q applications, is simply p + q applications.
Three.IV.2.24 Although these can be done by going through the indices, they are best understood
in terms of the represented maps. That is, fix spaces and bases so that the matrices represent linear
maps f, g, h.