Page 422 - 35Linear Algebra

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A 4 × 4 Gram Schmidt Example

4
Lets do an example of how to "Gram-Schmidt" some vectors in R . Given the
following vectors

       
o 0 3 1
1 1 0 1
       
v 1 =   , v 2 =   , v 3 =   , and v 4 =   ,
 0   1   1   0 
0 0 0 2

we start with v 1
 
0
1
 
⊥
v = v 1 =   .
1
 0 
0
Now the work begins
⊥
(v · v 2 ) ⊥
1
⊥
v 2 = v 2 − v 1
⊥ 2
kv k
1
   
0 0
1
1
  1  
=   −  
 1  1  0 
0 0
 
0
0
 
=  
 1 
0
This gets a little longer with every step.
⊥
⊥
(v · v 3 ) ⊥ (v · v 3 ) ⊥
⊥
1
2
v 3 = v 3 − v − v 2
1
⊥ 2
⊥ 2
kv k kv k
1 2
       
3 0 0 3
0
1
0
0
  0   1    
=   −   −   =  
1 1
 1   0   1   0 
0 0 0 0
u·v
This last step requires subtracting off the term of the form u for each of
u·u
the previously defined basis vectors.
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