Page 266 - 35Linear Algebra

266 Orthonormal Bases and Complements

Example 136 Find the QR decomposition of

 
2 −1 1
M =  1 3 −2  .
0 1 −2

What we will do is to think of the columns of M as three 3-vectors and use Gram–
Schmidt to build an orthonormal basis from these that will become the columns of
the orthogonal matrix Q. We will use the matrix R to record the steps of the Gram–
Schmidt procedure in such a way that the product QR equals M.
To begin with we write
 7   1 
2 − 1 1 0
5 5
M = 1 14 −2 0 1 0 .


 
5
0 1 −2 0 0 1
In the first matrix the first two columns are orthogonal because we simply replaced the
second column of M by the vector that the Gram–Schmidt procedure produces from
the first two columns of M, namely

 7     
− −1 2
5
 =  3 − 1 .
 14    1  
 5 5
1 1 0
1
The matrix on the right is almost the identity matrix, save the + in the second entry
5
of the first row, whose effect upon multiplying the two matrices precisely undoes what
we we did to the second column of the first matrix.
For the third column of M we use Gram–Schmidt to deduce the third orthogonal
vector
 1       7 
− 1 2 −
6 5
 = −2 − 0 1 −
 ,
 1      −9  14 
 3 54  5
− 7 −2 0 5 1
6
and therefore, using exactly the same procedure write
 7 1   1 
2 − − 1 0
5 6 5
 0
M = 1 14 1   1 −  .

5 
5 3 6
0 1 − 7 0 0 1
6
This is not quite the answer because the first matrix is now made of mutually orthog-
onal column vectors, but a bona fide orthogonal matrix is comprised of orthonormal
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