Re: symmetric SVD A=QSQ^T with Q orthonormal?
- From: Robert Israel <israel@xxxxxxxxxxx>
- Date: 16 May 2007 07:00:49 -0700
On May 16, 2:35 am, none <foont...@xxxxxxxxx> wrote:
Robert Israel wrote:
On May 15, 4:49 pm, none <foont...@xxxxxxxxx> wrote:
Robert Israel wrote:
none <foont...@xxxxxxxxx> writes:
Robert Israel wrote:
none <foont...@xxxxxxxxx> writes:
I'm familiar with Takagi factorizations for symmetric matrices
(symmetric
SVD). However, it can give complex eigenvectors.
I'd like to know if it is always possible to compute a "special"
SVD for a real symmetric matrices A such that:
A=QSQ^T
with Q orthonormal and S with singular values along the diagonal.
Intuitively it seems unlikely to me, but intuition should be used
to ask questions, not answer them.
If it can't be done always, can it be done if A is real, symmetric,
and nonnegative?
If A = Q S Q^T with Q an orthogonal matrix and S diagonal, the
diagonal entries of S are eigenvalues of A; the columns of Q are
eigenvectors. For a real symmetric matrix A, the singular values are
the absolute values
of the eigenvalues of A.
Why doesn't the following work?
Assume A is symmetric and not necessarily positive definite
Ax = lx
A^T Ax = A Ax = A lx = l^2 x
so x is also an eigenvector of A^T A, but with corresponding
eigenvalue l^2.
Since singular values are the positive square roots of eigenvalues of
A^T A,
and also because the singular vectors are the eigenvectors of A^T A,
shouldn't we be able to combine this to produce an SVD of the form:
A = Q S Q'
with Q containing the eigenvectors of A and S containing the absolute
values
of the eigenvalues of A?
We can.
If Q is a square matrix whose columns are orthonormal eigenvectors of
A, then A = Q D Q^T where D is a diagonal matrix whose diagonal entries
are the eigenvalues.
Now D = S U where S and U are diagonal, S_{i,i} = |D_{i,i}| and
U_{i,i} = (+/-) 1. The SVD is then A = Q S Q' with Q' = U Q^T.
I totally agree, but your method doesn't have Q' = Q^T right? (my
apologies for the misunderstanding due to my mistake with shoddy
notation)
In essence, I'm wondering why my "proof" does/doesn't show that there is
a Q such that
A = Q S Q^T
with Q orthogonal and S = diag(|D(i,i)|) when A is symmetric.
I guess what I don't understand is why you think there would be such a
Q.
If Q is an orthogonal matrix whose columns are eigenvectors of A,
then
Q^T A Q = D is a diagonal matrix whose diagonal entries are the
eigenvalues
of A. Not the singular values.
I guess my "proof" above has been misleading me for some time. I thought
that it showed that in the symmetric case there always exists a Q such that
D contained the singular values.
When we look at A^2 (A symmetric) then the positive square roots of its
eigenvalues are the singular values of A. Right?
Since the eigenvectors don't move after powering up A, I keep thinking I can
just use the same orthogonal Q and arrange the singular values in
descending order on the diagonal (moving the eigenvectors that correspond
to them too) to obtain the SVD.
A = Q S Q^T
Where is the flaw in that logic? I have a funny feeling I'm missing
something obvious.
The flaw is that Q S Q^T is not A. It's the positive semidefinite
square root of
A^2. If A isn't positive semidefinite, it's not A.
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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