Page 271 - 35Linear Algebra

14.6 Orthogonal Complements 271

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Example 139 Consider any line L through the origin in R . Then L is a subspace,
⊥
and L is a 3-dimensional subspace orthogonal to L. For example, let
 
 1 
 
1
 
 
L = span  
 
 1 
 
1
 
4
be a line in R . Then
  
x
 
 
y
   
L ⊥ =   ∈ R 4 (x, y, z, w) (1, 1, 1, 1) = 0
z
 
 
 
 
w

 
x
 
 
y
 
 
=   ∈ R 4 x + y + z + w = 0 .
z
 
 
 
w
 
⊥
Using the Gram-Schmidt procedure one may find an orthogonal basis for L . The set
     
1 1
 1 
 
 
 −1   0   0 
,
,
     
0 −1
     
 0 
 
0 0 −1
 
⊥
forms a basis for L so, first, we order the basis as
     
1 1 1
 −1   0   0 
,
(v 1 , v 2 , v 2 ) =       .
,
0 −1 0
     
0 0 −1
⊥
Next, we set v = v 1 . Then
1
     1 
1 1 2
v ⊥ =  0  − 1  −1  =  1 ,

2




2
 −1  2  0   −1 
0 0 0
 
     1  1
1 1 2 3
3
 .
v ⊥ =  0  − 1  −1  − 1/2  1  1 




2



3 = 
 0  2  0  3/2  −1   1 
3
−1 0 0 −1
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