Linear Algebra Exercises-n-Answers.pdf

54 Linear Algebra, by Hefferon
gives a linear system whose
⎛
solution
⎝ 1 3 x
⎞
⎛
2 2 y⎠ −2ρ1+ρ2 −2ρ 2+ρ 3
−→ −→ ⎝ 1 3 x
⎞
0 −4 −2x + y ⎠
−3ρ
3 1 z
1 +ρ 3
0 0 x − 2y + z
is possible if and only if the three-tall vector’s components x, y, and z satisfy x − 2y + z = 0. For
instance, we can find the coefficients c 1 and c 2 that work when x = 1, y = 1, and z = 1. However,
there are no c’s that work for x = 1, y = 1, and z = 2. Thus this is not a basis; it does not span the
space.
(c) Yes, this is a basis. Setting up the relationship leads to this reduction
⎛
⎝ 0 1 2 x
⎞
⎛
⎞
−1 1 0 z
2 1 5 y⎠ ρ1↔ρ3
−→ 2ρ1+ρ2
−→
−(1/3)ρ2+ρ3
−→ ⎝ 0 3 5 y + 2z ⎠
−1 1 0 z
0 0 1/3 x − y/3 − 2z/3
which has a unique solution for each triple of components x, y, and z.
(d) No, this is not a basis. The reduction
⎛
⎝ 0 1 1 x
⎞
⎛
2 1 3 y⎠ ρ 1↔ρ 3 2ρ 1 +ρ 2 (−1/3)ρ 2 +ρ 3
−→ −→ −→ ⎝ −1 1 0 z
⎞
0 3 3 y + 2z ⎠
−1 1 0 z
0 0 0 x − y/3 − 2z/3
which does not have a solution for each triple x, y, and z. Instead, the span of the given set includes
only those three-tall vectors where x = y/3 + 2z/3.
Two.III.1.17
(a) We solve
)
( ) (
−1 1
+ c 2 =
1 2)
(
1
c 1
1
with ( ) ( )
1 −1 1 −ρ 1+ρ 2 1 −1 1
−→
1 1 2 0 2 1
and conclude that c 2 = 1/2 and so c 1 = 3/2. Thus, the representation is this.
( ( )
1 3/2
Rep B ( ) =
2)
1/2
B
(b) The relationship c 1 · (1) + c 2 · (1 + x) + c 3 · (1 + x + x 2 ) + c 4 · (1 + x + x 2 + x 3 ) = x 2 + x 3 is easily
solved by eye to give that c 4 = 1, c 3 = 0, c 2 = −1, and c 1 = 0.
⎛ ⎞ ⎛ ⎞
0 0
(c) Rep E4
( ⎜−1
⎟
⎝ 0 ⎠ ) = ⎜−1
⎟
⎝ 0 ⎠
1 1
E 4
⎛ ⎞
0
Rep D (x 2 + x 3 ) = ⎜−1
⎟
⎝ 0 ⎠
1
Two.III.1.18 A natural basis is 〈1, x, x 2 〉. There are bases for P 2 that do not contain any polynomials
of degree one or degree zero. One is 〈1 + x + x 2 , x + x 2 , x 2 〉. (Every basis has at least one polynomial
of degree two, though.)
Two.III.1.19 The reduction ( ) ( )
1 −4 3 −1 0 −2ρ 1 +ρ 2 1 −4 3 −1 0
−→
2 −8 6 −2 0
0 0 0 0 0
gives that
⎛
the only condition
⎞
is that x 1 = 4x 2 − 3x
⎛ 3 +
⎞
x 4 . The
⎛
solution
⎞
set
⎛
is
⎞
4x 2 − 3x 3 + x 4
4 −3 1
{ ⎜ x 2
⎟ ∣
⎝ x 3
⎠ x 2 , x 3 , x 4 ∈ R} = {x 2
⎜1
⎟
⎝0⎠ + x ⎜ 0
⎟
3 ⎝ 1 ⎠ + x ⎜0
⎟ ∣
4 ⎝0⎠
x 2 , x 3 , x 4 ∈ R}
x 4
0 0 1
and so the obvious candidate for the basis ⎛is⎞
this. ⎛ ⎞ ⎛ ⎞
4 −3 1
〈 ⎜1
⎟
⎝0⎠ , ⎜ 0
⎟
⎝ 1 ⎠ , ⎜0
⎟
⎝0⎠ 〉
0 0 1
We’ve shown that this spans the space, and showing it is also linearly independent is routine.
D