A First Course in Linear Algebra, 2017a

4.11. Orthogonality and the Gram Schmidt Process 245
Solution. We will first use the Gram-Schmidt Process to construct the orthogonal basis, B,ofW :
⎧⎡
⎤ ⎡ ⎤ ⎡ ⎤⎫
1 0 1
⎪⎨
B = ⎢ 0
⎥
⎣ 1 ⎦
⎪⎩
, ⎢ 0
⎥
⎣ 0 ⎦ , ⎢ 2
⎪⎬
⎥
⎣ −1 ⎦ .
⎪⎭
0 1 0
By Theorem 4.146,
proj W (⃗y)= 2 2
⎡
⎢
⎣
1
0
1
0
⎤
⎥
⎦ + 5 1
⎡
⎢
⎣
0
0
0
1
⎤
⎥
⎦ + 12
6
⎡
⎢
⎣
1
2
−1
0
⎤
⎡
⎥
⎦ = ⎢
⎣
3
4
−1
5
⎤
⎥
⎦
is the vector in W closest to⃗y.
♠
Consider the next example.
Example 4.148: Vector Written as a Sum of Two Vectors
⎧⎡
⎤ ⎡ ⎤⎫
1 0
⎪⎨
Let W be a subspace given by W = span ⎢ 0
⎥
⎣ 1 ⎦
⎪⎩
, ⎢ 1
⎪⎬
⎥
⎣ 0 ⎦ ,andY =(1,2,3,4).
⎪⎭
0 2
Find the point Z in W closest to Y , and moreover write⃗y as the sum of a vector in W and a vector in
W ⊥ .
Solution. From Theorem 4.139, the point Z in W closest to Y is given by⃗z = proj W (⃗y).
Notice that since the above vectors already give an orthogonal basis for W, wehave:
⃗z = proj W (⃗y)
( ) ( )
⃗y •⃗w1 ⃗y •⃗w2
=
‖⃗w 1 ‖ 2 ⃗w 1 +
‖⃗w 2 ‖ 2 ⃗w 2
⎡ ⎤ ⎡ ⎤
(
1
( )
0
4 = ⎢ 0
⎥ 10
2)
⎣ 1 ⎦ + ⎢ 1
⎥
5 ⎣ 0 ⎦
0
2
Therefore the point in W closest to Y is Z =(2,2,2,4).
=
⎡
⎢
⎣
2
2
2
4
⎤
⎥
⎦