Page 167 - Linear Algebra Done Right

Polar and Singular-Value Decompositions
The polar decomposition (7.41) states that each operator on V is the
product of an isometry and a positive operator. Thus we can write each
operator on V as the product of two operators, each of which comes 155
from a class that we have completely described and that we under-
stand reasonably well. The isometries are described by 7.37 and 7.38;
the positive operators (which are all self-adjoint) are described by the
spectral theorem (7.9 and 7.13).
√
Specifically, suppose T = S T T is the polar decomposition of
∗
T ∈L(V), where S is an isometry. Then there is an orthonormal basis
of V with respect to which S has a diagonal matrix (if F = C) or a block
diagonal matrix with blocks of size at most 2-by-2 (if F = R), and there
√
is an orthonormal basis of V with respect to which T T has a diag-
∗
onal matrix. Warning: there may not exist an orthonormal basis that
√
simultaneously puts the matrices of both S and T T into these nice
∗
forms (diagonal or block diagonal with small blocks). In other words, S
√
may require one orthonormal basis and T T may require a different
∗
orthonormal basis.
Suppose T ∈L(V). The singular values of T are the eigenvalues
√ √
of T T, with each eigenvalue λ repeated dim null( T T − λI) times.
∗
∗
The singular values of T are all nonnegative because they are the eigen-
√
values of the positive operator T T.
∗
4
For example, if T ∈L(F ) is defined by
7.45 T(z 1 ,z 2 ,z 3 ,z 4 ) = (0, 3z 1 , 2z 2 , −3z 4 ),
then T T(z 1 ,z 2 ,z 3 ,z 4 ) = (9z 1 , 4z 2 , 0, 9z 4 ), as you should verify. Thus
∗
√
T T(z 1 ,z 2 ,z 3 ,z 4 ) = (3z 1 , 2z 2 , 0, 3z 4 ),
∗
√
and we see that the eigenvalues of T T are 3, 2, 0. Clearly
∗
√ √ √
dim null( T T −3I) = 2, dim null( T T −2I) = 1, dim null T T = 1.
∗
∗
∗
Hence the singular values of T are 3, 3, 2, 0. In this example −3 and 0
are the only eigenvalues of T, as you should verify.
Each T ∈L(V) has dim V singular values, as can be seen by applying
the spectral theorem and 5.21 (see especially part (e)) to the positive
√
(hence self-adjoint) operator T T. For example, the operator T de-
∗
4
fined by 7.45 on the four-dimensional vector space F has four singular
values (they are 3, 3, 2, 0), as we saw in the previous paragraph.
The next result shows that every operator on V has a nice descrip-
tion in terms of its singular values and two orthonormal bases of V.