Dear allstat users,
here there is a summary about the query reported at the bottom.
I have to thank Prof. Alistair Watson for his help.
Case of M real symmetric
Then there is an orthogonal matrix Q such that
Q^T M Q = D,
where D is a diagonal matrix containing the eigenvalues.
For the SVD of M, there are orthogonal matrices U, V such that
M = U^T S V
where
S contains the singular values (these are the moduli of the elements of D);
the matrices U, V are orthogonal (U and V may differ by signs of columns).
Case of M real but not necessarily symmetric
In this case the eigenvalues of M may not be real. There may not be an orthogonal matrix of eigenvectors.
All one can say is that if the SVD is
M= U S V^T
then
M^TM = V S^2 V^T
that is the eigenvalues of M^TM are the squares of the singular values of M, and the matrix of eigenvectors of M^TM is V.
ORIGINAL QUESTION:
> Dear allstat users,
> here there is a question about matrix algebra.
>
> Given a squared matrix M, according to eigenvalues theory (or eigenvalue decomposition), I can find a matrix K such that J = KMK^{T} is diagonal, with K^{T} being the transpose of K.
>
> Another important factorisation for M is called Singular Value Decomposition (SVD). In this case M = USV^{*} where
> U is a unitary matrix, S is a matrix with nonnegative numbers on the diagonal and zero off the diagonal and V^{*} denotes the conjugate transpose of V, a unitary matrix.
>
> I think that in some cases, and in particular case of M being Hermitian, these two factorisations are strictly related.
> Could anyone explain to me briefly which is the relation between these two decompositions?
>
> thank you
> Stefano Sofia PhD
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