A First Course in Linear Algebra, 2017a

4.11. Orthogonality and the Gram Schmidt Process 241
Theorem 4.139: Approximation Theorem
Let W be a subspace of R n and Y any point in R n .LetZ be the point whose position vector is the
orthogonal projection of Y onto W .
Then, Z is the point in W closest to Y .
Proof. First Z is certainly a point in W since it is in the span of a basis of W.
To show that Z is the point in W closest to Y , we wish to show that |⃗y −⃗z 1 | > |⃗y−⃗z| for all⃗z 1 ≠⃗z ∈ W.
We begin by writing ⃗y −⃗z 1 =(⃗y −⃗z)+(⃗z −⃗z 1 ). Now, the vector ⃗y −⃗z is orthogonal to W, and⃗z −⃗z 1 is
contained in W. Therefore these vectors are orthogonal to each other. By the Pythagorean Theorem, we
have that
‖⃗y −⃗z 1 ‖ 2 = ‖⃗y −⃗z‖ 2 + ‖⃗z −⃗z 1 ‖ 2 > ‖⃗y −⃗z‖ 2
This follows because⃗z ≠⃗z 1 so ‖⃗z −⃗z 1 ‖ 2 > 0.
Hence, ‖⃗y −⃗z 1 ‖ 2 > ‖⃗y −⃗z‖ 2 . Taking the square root of each side, we obtain the desired result.
Consider the following example.
Example 4.140: Orthogonal Projection
Let W be the plane through the origin given by the equation x − 2y + z = 0.
Find the point in W closest to the point Y =(1,0,3).
♠
Solution. We must first find an orthogonal basis for W. Notice that W is characterized by all points (a,b,c)
where c = 2b − a. Inotherwords,
⎡
a
⎤ ⎡
1
⎤ ⎡ ⎤
0
W = ⎣ b
2b − a
⎦ = a⎣
0
−1
⎦ + b⎣
1 ⎦, a,b ∈ R
2
We can thus write W as
W = span{⃗u 1 ,⃗u 2 }
⎧⎡
⎤ ⎡
⎨ 1
= span ⎣ 0 ⎦, ⎣
⎩
−1
Notice that this span is a basis of W as it is linearly independent. We will use the Gram-Schmidt
Process to convert this to an orthogonal basis, {⃗w 1 ,⃗w 2 }. In this case, as we remarked it is only necessary
to find an orthogonal basis, and it is not required that it be orthonormal.
⃗w 2
⎡
⃗w 1 = ⃗u 1 = ⎣
1
0
−1
⎤
⎦
( ) ⃗u2 •⃗w 1
= ⃗u 2 −
‖⃗w 1 ‖ 2 ⃗w 1
0
1
2
⎤⎫
⎬
⎦
⎭