Linear Algebra, 2020a

Section VI. Projection 291
is perpendicular to each member of the basis so
⃗0 = A T( ⃗v − A⃗c ) = A T ⃗v − A T A⃗c
and solving gives this (showing that A T A is invertible is an exercise).
⃗c = ( A T A ) −1
AT · ⃗v
Therefore proj M (⃗v )=A · ⃗c = A(A T A) −1 A T · ⃗v, as required.
3.9 Example To orthogonally project this vector into this subspace
⎛ ⎞ ⎛ ⎞
1
x
⎜ ⎟ ⎜ ⎟
⃗v = ⎝−1⎠ P = { ⎝y⎠ | x + z = 0}
1
z
first make a matrix whose columns are a basis for the subspace
⎛ ⎞
0 1
⎜ ⎟
A = ⎝1 0 ⎠
0 −1
and then compute.
⎛
A ( A T A ) −1
A T ⎜
0 1
⎞
( )( )
⎟ 1 0 0 1 0
= ⎝1 0 ⎠
0 1/2 1 0 −1
0 −1
⎛
⎞
1/2 0 −1/2
⎜
⎟
= ⎝ 0 1 0 ⎠
−1/2 0 1/2
QED
With the matrix, calculating the orthogonal projection of any vector into P is
easy.
⎛
⎞ ⎛ ⎞ ⎛ ⎞
1/2 0 −1/2 1 0
⎜
⎟ ⎜ ⎟ ⎜ ⎟
proj P (⃗v) = ⎝ 0 1 0 ⎠ ⎝−1⎠ = ⎝−1⎠
−1/2 0 1/2 1 0
Note, as a check, that this result is indeed in P.
Exercises
̌ 3.10 Project
( )
the vectors
(
into M along N.
( 3
x x
(a) , M = { | x + y = 0}, N = { | −x − 2y = 0}
−2
y)
y)
( ( ( 1 x x
(b) , M = { | x − y = 0}, N = { | 2x + y = 0}
2)
y)
y)
⎛ ⎞
3
(c) ⎝0⎠ ,
⎛ ⎞
x
M = { ⎝y⎠ | x + y = 0},
⎛ ⎞
1
N = {c · ⎝0⎠ | c ∈ R}
1
z
1
̌ 3.11 Find M ⊥ .