Re: symmetric SVD A=QSQ^T with Q orthonormal?

Robert Israel wrote:

none <foontala@xxxxxxxxx> writes:

Robert Israel wrote:

none <foontala@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.

Thanks for your input, I really appreciate your thoughts.


.

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