Eigenvalues of a complex symmetric matrix

[sodaboy7]

Eigen values of a complex symmetric matrix which is NOT a hermitian are not always real. I want to formulate conditions for which eigen values of a complex symmetric matrix (which is not hermitian) are real.

 

[Robert1986]

Well, assume $A$ has real eigenvalues. Then if $\lambda,x$ are an "eigenpair" we have $x^*Ax=x^*\lambda x = \lambda x^*x$ which is real. On the other hand, $\lambda x^* x = (\lambda x)^*x = (Ax)^*x = x^*A^*x$ so that we have $x^*Ax = x^*A^*x$. Now, if $A$ is symmetric, I *think* this means it can be diagonalised (ie there is an eigenbasis) and so this argument seems like it might imply that $A=A^*$, that is, $A$ is Hermitian. However, I don't have a lot of time to think about it now, and I might be missing something important.

 

[sodaboy7]

Do you mean if a complex symmetric matrix is diagonizable, it will have real eigenvalues?

 

[Robert1986]

No, I *think* all symmetric matrices are diagonalisable (and thus have an eigenbasis) and if all eigenvalues are real, then the matrix is Hermitian. That is, I am saying that a symmetric matrix is hermitian iff all eigenvalues are real.

 

[AlephZero]

That is, I am saying that a symmetric matrix is hermitian iff all eigenvalues are real.

A symmetric matrix is hermitian iff the matrix is real, so that is not a good way to characterize symmetric complex matrices.

I don't think there is a simple answer to the OP's question.

 

[Robert1986]

A symmetric matrix is hermitian iff the matrix is real, so that is not a good way to characterize symmetric complex matrices.

I don't think there is a simple answer to the OP's question.

I should have been more clear. Any symmetric matrix $M$ has an eigenbasis (because any symmetric matrix is diagonalisable.) Now, if all the eigenvalues of a symmetric matrix are real, then $A^* = A$, ie, $A$ is hermitian. However, as you pointed out, since $A^\top = A$ by assumption, this implies that $A$ is real.

So, what I am saying is that there are no complex symmetric matrices with all real eigenvalues. (Unless, of course, the matrix is real.) In other words, given that $M$ is symmetric and has real eigenvalues, then it must be real.

EDIT:
In fact, given any matrix $M$, if $x^*M^*x$ is real for all $x$ then $M$ is hermitian.

 

[AlephZero]

So, what I am saying is that there are no complex symmetric matrices with all real eigenvalues. (Unless, of course, the matrix is real.)

OK, I agree with that.

But the number of complex eigenvalues can by anything from 1 to the order of the matrix, which doesn't go very far to answer the OP's question.

 

[Robert1986]

OK, I agree with that.

But the number of complex eigenvalues can by anything from 1 to the order of the matrix, which doesn't go very far to answer the OP's question.

Perhaps I misunderstood the OP, but it seems that he wants to know under what conditions the eigenvalues of a symmetric matrix are real. The answer is the the matrix must be real.

 

[AlephZero]

Maybe we both misunderstood, but I read the OP's "which eigen values ... are real" as a question about some of them, not all of them.

It might be possible to give an answer if the eigenproblem represents a physical system. For example the eigenvalues of a damped multi-degree-of-freedom oscillator, with an arbitrary damping matrix, represent the damped natural freuqencies on the s-plane, therefore they are all complex except for zero-frequency (rigid body motion) modes. Also, the sign of the real part of the eigenvalues shows whether the mode is damped, undamped, or unstable (i.e. it gains energy from outside the system).

But I don't know how to turn that "physics insight" about a particular physical system into a mathematical way to characaterize the matrix.

 

[Robert1986]

Maybe we both misunderstood, but I read the OP's "which eigen values ... are real" as a question about some of them, not all of them.

It might be possible to give an answer if the eigenproblem represents a physical system. For example the eigenvalues of a damped multi-degree-of-freedom oscillator, with an arbitrary damping matrix, represent the damped natural freuqencies on the s-plane, therefore they are all complex except for zero-frequency (rigid body motion) modes. Also, the sign of the real part of the eigenvalues shows whether the mode is damped, undamped, or unstable (i.e. it gains energy from outside the system).

But I don't know how to turn that "physics insight" about a particular physical system into a mathematical way to characaterize the matrix.

Ha, I see now. It seems we are interpreting things differently. Looking back at the OP, it seems that your interpretation is more in line with what the OP wrote; however I prefer my interpretation because it admits a solution :).

OK, given your interpretation (which I now think is the correct one), I agree that the problem is hard and not very well formulated. "Which eigenvalues are real?" is kind of an odd one to answer. I mean, "which" in what sense?

 

[sodaboy7]

What I am trying to say is this.
All hermitian matrices are symmetric but all symmetric matrices are not hermitian. Eigenvalues of hermitian (real or complex) matrices are always real. But what if the matrix is complex and symmetric but not hermitian. In hermitian the ij element is complex conjugal of ji element. But I am taking about matrix for which ij element and ji element are equal. Eigen values of such a matrix may not be real. So under what condition Eigenvalues will be real.

 

[Robert1986]

What I am trying to say is this.
All hermitian matrices are symmetric but all symmetric matrices are not hermitian. Eigenvalues of hermitian (real or complex) matrices are always real. But what if the matrix is complex and symmetric but not hermitian. In hermitian the ij element is complex conjugal of ji element. But I am taking about matrix for which ij element and ji element are equal. Eigen values of such a matrix may not be real. So under what condition Eigenvalues will be real.

First of all, a hermitian matrix is symmetric if and only if the matrix is real. A hermitian complex matrix is not symmetric.

But, to answer your question, the matrix must be real. That is, if a matrix is symmetric and has real eigenvalues, then it is a real matrix. Does this make sense? Put another way, all symmetric matrices with real eigenvalues are real matrices.

 

[Erland]

I should have been more clear. Any symmetric matrix $M$ has an eigenbasis (because any symmetric matrix is diagonalisable.)

But it has not been proved in this thread (nor is a reference to a proof given) that a symmetric matrix must be diagonalizable.

EDIT:
In fact, given any matrix $M$, if $x^*M^*x$ is real for all $x$ then $M$ is hermitian.

But it is not proved, at this stage, that $x^*Mx$ is real for all $x$, even if $M$ is symmetric and diagonalizable with all eigenvalues real. In that case, this holds if $x$ is an eigenvector to $M$, but since we don't know that the eigenbasis is orthogonal, this cannot be generalized to all $x$.

 

[Robert1986]

But it has not been proved in this thread (nor is a reference to a proof given) that a symmetric matrix must be diagonalizable.

Yes, and as I think about it, there are really simple counterexamples to what I said.

So, forget what I wrote...

 

[Robert1986]

However, as proven in Matrix Analysis, if a symmetric matrix is diagonalisable, then it is diagonalisable via an orthogonal matrix, and so what I wrote does follow. NOW, if the matrix is not diagonalisable, there is obviously not an eigenbasis (orthogonal or otherwise.)

Now, if the matrix is normal (commutes with its adjoint) then it is orthogonally diagonalisable (this is also in Matrix Analysis), and what I wrote then follows.

EDIT:
Again, from matrix analysis, if $M$ is complex-symmetric, there is a unitary matrix $U$ such that $M=UDU^\top$ where the columns of $U$ are eigenvectors of $MM^* = M\bar{M}$ and $D$ is diagonal and the entries are the positive square roots of the corresponding eigenvalues. Now, if the columns of $U$ are real, the $U$ is orthogonal and so $M$ is orthogonally diagonalisable, and what I wrote follows. So, IF the eigenvectors of $M\bar{M}$ are real, then the eigenvalues of $M$ are real.