Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 139
(g) Here, M is one-dimensional
⎛
M = {c · ⎝ 0 ⎞
⎛
−1⎠ ∣ c ∈ R} B M = 〈 ⎝ 0 ⎞
−1⎠〉
1
1
and as a result, M ⊥ is two-dimensional.
⎛ ⎞
M ⊥ = { ⎝ u v ⎠ ∣ 0 · u − 1 · v + 1 · w = 0} = {j · ⎝ 1 0⎠ + k · ⎝ 0 1⎠ ∣ j, k ∈ R}
w
0 1
Three.VI.3.12 (a) Paramatrizing the equation leads to this basis for P .
⎛ ⎞ ⎛ ⎞
B P = 〈 ⎝ 1 0⎠ , ⎝ 0 1⎠〉
3 2
(b) Because R 3 is three-dimensional and P is two-dimensional, the complement P ⊥ must be a line.
Anyway, the calculation as in Example 3.5
⎛
P ⊥ = { ⎝ x ⎞
y⎠ ∣ ( ) ⎛ 1 0 3 ⎝ x ⎞
(
y⎠ 0
= }
0 1 2
0)
z
z
gives this basis for P ⊥ .
⎛
⎞
⎛
⎞
(c) ⎝ 1 1⎠ = (5/14) · ⎝ 1 0⎠ + (8/14) · ⎝ 0 1⎠ + (3/14) ·
2
⎛
3
2
(d) proj P ( ⎝ 1 ⎞ ⎛
1⎠) = ⎝ 5/14 ⎞
8/14 ⎠
2 31/14
(e) The matrix of the projection
⎛
⎝ 1 0
⎞
0 1⎠ (( ) ⎛ 1 0 3
⎝ 1 0
⎞
0 1⎠ ) −1
0 1 2
3 2
3 2
⎛
⎞
B P ⊥ = 〈 ⎝ 3 2 ⎠〉
−1
⎛ ⎞ ⎛ ⎞
⎝ 3 2
−1
⎠
⎛
( ) 1 0 3
=
0 1 2
when applied to the vector, yields the expected result.
⎛
1
⎝ 5 −6 3
⎞ ⎛
−6 10 2 ⎠ ⎝ 1 ⎞
1⎠ =
14
3 2 13 2
Three.VI.3.13
(a) Paramatrizing gives this.
M = {c ·
⎛
( ) −1 ∣∣
c ∈ R}
1
⎞
⎛
⎞
⎝ 1 0
⎞
( ) −1 ( )
0 1⎠
10 6 1 0 3
6 5 0 1 2
3 2
⎛
= 1 ⎝ 5 −6 3
⎞
−6 10 2 ⎠
14
3 2 13
⎛
⎝ 5/14
⎞
8/14 ⎠
31/14
For the first way, we take the vector spanning the line M to be
( ) −1
⃗s =
1
and the Definition 1.1 formula gives this.
( ) ( )
1 −1
( )
1 −3 1
proj [⃗s ] ( ) = ( ) ( ) ·
−3 −1 −1
1 1
For the second way, we fix
( )
−1
B M = 〈 〉
1
( )
−1
= −4
1 2 ·
( ) (
−1 2
=
1 −2)