Linear Algebra Exercises-n-Answers.pdf

58 Linear Algebra, by Hefferon
and so a natural candidate for a basis is this. ⎛ ⎞ ⎛ ⎞
〈 ⎝ 0 1⎠ , ⎝ 0 0⎠〉
0 1
To check linear independence we set up
⎛
c 1
⎝ 0 ⎞ ⎛
1⎠ + c 2
⎝ 0 ⎞ ⎛
0⎠ = ⎝ 1 ⎞
0⎠
0 1 0
(the vector on the right is the zero object in this space). That yields the linear system
(−c 1 + 1) + (−c 2 + 1) − 1 = 1
c 1 = 0
c 2 = 0
with only the solution c 1 = 0 and c 2 = 0. Checking the span is similar.
Subsection Two.III.2: Dimension
Two.III.2.14
One basis is 〈1, x, x 2 〉, and so the dimension is three.
Two.III.2.15 The solution set is ⎛
⎞
4x 2 − 3x 3 + x 4
{ ⎜ x 2
⎟ ∣
⎝ x 3
⎠ x 2 , x 3 , x 4 ∈ R}
x 4
so a natural basis is this
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
4 −3 1
〈 ⎜1
⎟
⎝0⎠ , ⎜ 0
⎟
⎝ 1 ⎠ , ⎜0
⎟
⎝0⎠ 〉
0 0 1
(checking linear independence is easy). Thus the dimension is three.
Two.III.2.16 For
(
this
)
space
( )
a b ∣∣ 1 0
{ a, b, c, d ∈ R} = {a · + · · · + d ·
c d
0 0
this is a natural basis. ( ) ( ) ( )
1 0 0 1 0 0
〈 , , ,
0 0 0 0 1 0
The dimension is four.
( )
0 0 ∣∣
a, b, c, d ∈ R}
0 1
( )
0 0
〉
0 1
Two.III.2.17 (a) As in the prior exercise, the space M 2×2 of matrices without restriction has this
basis
( ) ( ) ( ) ( )
1 0 0 1 0 0 0 0
〈 , , , 〉
0 0 0 0 1 0 0 1
and so the dimension is four.
(b) For this space
( ) ( ) ( ) ( )
a b ∣∣ 1 1 −2 0 0 0 ∣∣
{ a = b − 2c and d ∈ R} = {b · + c · + d · b, c, d ∈ R}
c d
0 0 1 0 0 1
this is a natural basis.
( ) ( ) ( )
1 1 −2 0 0 0
〈 , , 〉
0 0 1 0 0 1
The dimension is three.
(c) Gauss’ method applied to the two-equation linear system gives that c = 0 and that a = −b.
Thus, we have this
(
description
) ( ) ( ) −b b ∣∣ −1 1 0 0 ∣∣
{
b, d ∈ R} = {b · + d · b, d ∈ R}
0 d
0 0 0 1
and so this is a natural basis. ( ) ( )
−1 1 0 0
〈 , 〉
0 0 0 1
The dimension is two.