Page 311 - 35Linear Algebra

17.2 Singular Value Decomposition 311

are eigenvectors of
1 0 − 1
 
2 2
T
MM =  0 2 0 
− 1 0 1
2 2
with eigenvalues 1 and 2, respectively. The third eigenvector (with eigenvalue 0) of
MM T is
 1 
√
2
v 3 =  0 .


√ 1
2
The eigenvectors Mu 1 and Mu 2 are necessarily orthogonal, dividing them by their
lengths we obtain the left singular vectors and in turn our orthonormal output basis

 1     1 
√ 0 √
2 2
0      
0  ,  −1  ,  0 .
O = 
− √ 1 0 √ 1
2 2
0
The new matrix M of the linear transformation given by M with respect to the bases
0
O and O is
 
1 √ 0
0
M =  0 2  ,
0 0
√
so the singular values are 1, 2.
0
Finally note that arranging the column vectors of O and O into change of basis
matrices
 1 1 
√ 0 √
1 1 ! 2 2
√ √
2 2  
P = , Q =  0 −1 0  ,
√ 1 − √ 1  
2 2 1 1
− √ 0 √
2 2
we have, as usual,
0
M = Q −1 MP .

Singular vectors and values have a very nice geometric interpretation;
they provide an orthonormal bases for the domain and range of L and give
the factors by which L stretches the orthonormal input basis vectors. This
is depicted below for the example we just computed.

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