A First Course in Linear Algebra, 2017a

244 R n
In order to find W ⊥ , we need to find all⃗x which are orthogonal to every vector in this span.
⎡ ⎤
x 1
Let⃗x = ⎣ x 2
⎦. In order to satisfy⃗x •⃗u 1 = 0, the following equation must hold.
x 3
x 1 − x 3 = 0
In order to satisfy⃗x •⃗u 2 = 0, the following equation must hold.
x 2 + 2x 3 = 0
Both of these equations must be satisfied, so we have the following system of equations.
x 1 − x 3 = 0
x 2 + 2x 3 = 0
To solve, set up the augmented matrix.
[ 1 0 −1 0
]
0 1 2 0
⎧⎡
⎨
Using Gaussian Elimination, we find that W ⊥ = span ⎣
⎩
for W ⊥ .
1
−2
1
⎤⎫
⎧⎡
⎬ ⎨
⎦
⎭ , and hence ⎣
⎩
The following results summarize the important properties of the orthogonal projection.
Theorem 4.146: Orthogonal Projection
1
−2
1
⎤⎫
⎬
⎦ is a basis
⎭
Let W be a subspace of R n , Y be any point in R n ,andletZ be the point in W closest to Y . Then,
1. The position vector⃗z of the point Z is given by⃗z = proj W (⃗y)
2. ⃗z ∈ W and⃗y −⃗z ∈ W ⊥
3. |Y − Z| < |Y − Z 1 | for all Z 1 ≠ Z ∈ W
♠
Consider the following example of this concept.
Example 4.147: Find a Vector Closest to a Given Vector
Let
⃗x 1 =
⎡
⎢
⎣
1
0
1
0
⎤
⎥
⎦ ,⃗x 2 =
⎡
⎢
⎣
1
0
1
1
⎤
⎥
⎦ ,⃗x 3 =
We want to find the vector in W = span{⃗x 1 ,⃗x 2 ,⃗x 3 } closest to⃗y.
⎡
⎢
⎣
1
1
0
0
⎤ ⎡
⎥
⎦ , and⃗y = ⎢
⎣
4
3
−2
5
⎤
⎥
⎦ .