Page 428 - 35Linear Algebra

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Moreover, the vector
 
a
 0 
0
√ √
is the rotated v 1 so must have length ||v 1 || = 3. Thus a = 3.
The rotated v 2 is
 
b
 d 
0
√
and must have length ||v 2 || = 2 2. Also the dot product between
   
a b
 0  and  d 
0 0

is ab and must equal v 1 v 2 = 0. (That v 1 and v 2 were orthogonal is just a
coincidence here... .) Thus b = 0. So now we know most of the matrix R
√ 
3 √ 0 c
R =  0 2 2 e  .
0 0 f

You can work out the last column using the same ideas. Thus it only remains to
compute Q from
Q = MR −1 .

G.14 Diagonalizing Symmetric Matrices

3 × 3 Example

Lets diagonalize the matrix

 
1 2 0
M =  2 1 0 
0 0 5

If we want to diagonalize this matrix, we should be happy to see that it
is symmetric, since this means we will have real eigenvalues, which means
factoring won’t be too hard. As an added bonus if we have three distinct
eigenvalues the eigenvectors we find will automatically be orthogonal, which
means that the inverse of the matrix P will be easy to compute. We can start

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