Page 311 - 35Linear Algebra
P. 311
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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