Diagonalization of complex symmetric matrices

[petergreat]

Is every complex symmetric (NOT unitary) matrix $$M$$ diagonalizable in the form $$U^T M U$$, where $$U$$ is a unitary matrix? Why?

 

[g_edgar]

Woah, $$U^T$$ in your formula and not $$U^*$$ ... so in general $$U^T$$ is not the inverse of $$U$$ . Why did you choose that?

 

[petergreat]

Woah, $$U^T$$ in your formula and not $$U^*$$ ... so in general $$U^T$$ is not the inverse of $$U$$ . Why did you choose that?

Because I want to diagonalize a quadratic form $$v^T M v$$ where v is a complex vector. No complex conjugation is involved, so the only useful form of diagonalization is $$U^T M U$$.

I met this problem in physics. A specific complex symmetric matrix is involved, and it is diagonalized by an ansatz for the unitary matrix $$U$$. However, I want to know whether this can work in general.

P.S. I know that $$U^T$$ is not the inverse of $$U$$. Otherwise it would be too standard and I wouldn't need to ask here.
(In case anyone wants to know where this comes from, the mass term for Majorana neutrinos is essentially such a quadratic form; on the other hand, the mass term for Dirac neutrinos involve complex conjugation and can be dealt with in the usual manner of diagonalization $$U^\dagger M U$$)

 

[jostpuur]

Is every complex symmetric (NOT unitary) matrix $$M$$ diagonalizable in the form $$U^T M U$$, where $$U$$ is a unitary matrix? Why?

You probably meant to emphasize "symmetric (not hermitian)"?

My belief is that if $M\in\mathbb{C}^{n\times n}$ is symmetric so that $M^T=M$, then there exists a complex orthogonal matrix $O\in\mathbb{C}^{n\times n}$ so that $O^T=O^{-1}$, and so that $O^TMO$ is diagonal. (And I believe that the answer to your question is: No.)

Unfortunately I don't have a reference for this claim, and I also don't have energy to go through a proof right now, because this isn't my problem, so you shouldn't believe my belief

I would recommend finding out how to prove the well known diagonalizability results of real symmetric, and complex hermitian matrices, and see if you can modify the proofs.

This is a little bit paradoxical topic for me, because I never studied these proofs from any educational material, but at some point I felt like my understanding on linear algebra had grown to a point when I could prove these results on my own. According to my understanding the proofs can be carried out recursively by using the fact that any matrix will always have at least one eigenvalue, and then for the purpose of moving the smaller dimensional subspaces (dim)$n\mapsto n-1$ you use some invariance properties such as: Orthogonal transformation keeps symmetric matrix as symmetric, or unitary transformation keeps hermitian matrix as hermitian.

 

[petergreat]

You probably meant to emphasize "symmetric (not hermitian)"?

Yes, that was a typo.

My belief is that if $M\in\mathbb{C}^{n\times n}$ is symmetric so that $M^T=M$, then there exists a complex orthogonal matrix $O\in\mathbb{C}^{n\times n}$ so that $O^T=O^{-1}$, and so that $O^TMO$ is diagonal. (And I believe that the answer to your question is: No.)

Unfortunately I don't have a reference for this claim, and I also don't have energy to go through a proof right now, because this isn't my problem, so you shouldn't believe my belief

That was also what I thought. What you described is sufficient to diagonalize the quadratic form. But the authors wanted to diagonalize the quadratic form while stilling preserving orthonormality, so went for unitary matrices, unfortunately (to me) with success...

 

[jostpuur]

Can you show explicitly your example matrices that you have been working on?

 

[petergreat]

Can you show explicitly your example matrices that you have been working on?

I've attached a snapshot of my write-up of what I read. Basically the matrix contains a first-order small parameter, and is diagonalized by a unitary matrix that is expanded around identity up to 2nd order. The unitary matrix block-diagonalizes the matrix, and it is assumed that each block can be separately further diagonalized by unitary matrices. (The block-diagonalization is performed first to prove that some eigenvalues are an order smaller than the others. But this is not important to our discussion). If the assumption that each block can be separately further diagonalized is dubious, I've at least showed that 2 by 2 complex symmetric matrices of a particular form can be diagonalized in this manner.