If a N×N (N≥3) Hermitian matrix A meets the following conditions:

  1. A is positive semi-definite (not positive definite, i.e. A has at least M zero eigenvalue, where M is a given paremeter with 1≤M≤N-1).
  2. The sum of each off diagonal results in 0, and the main diagonal elements are non-negative, which is shown in the figure (set N=4 as example).

Then what the general solution of A is?

For example, a particular solution of A can be $$ \begin{matrix} I_{M'} & 0 &\\ 0 & 0 \\ \end{matrix} $$ where M≤N-M'≤N-1. It is just a particular solution, I wonder what is the general solution under these two conditions.

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closed as unclear what you're asking by Brian Borchers, Claude Leibovici, user228113, David K, Aqua Nov 28 '17 at 18:25

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  • $\begingroup$ Welcome to MSE. Here's some guide to MathJax. $\endgroup$ – Sou Nov 22 '17 at 12:49
  • $\begingroup$ In a question like this some examples using numbers would be nice. $\endgroup$ – P. Siehr Nov 22 '17 at 13:46
  • $\begingroup$ After the last edit, your question does not contain a question any more. Then what the general solution of A is? What does that mean? I voted to close this question for now until this question is formulated well, and can be answered. Right now, with edit after edit, it will only annoy everyone who already answered, as their solution will get incorrect. $\endgroup$ – P. Siehr Nov 23 '17 at 8:25

I claim that your matrices can never be decomposed in such a way.

First, note that a decomposition of the form $ A = BB^\dagger$ is possible in general if and only if your Hermitian matrix $A$ is positive semi-definite.

You seem to require that every element of the main diagonal be zero. We can in fact weaken this condition to requiring the matrices be trace-free. Since all the eigenvalues of a Hermitian matrix are real, it follows that there must be at least one positive eigenvalue and at least one negative eigenvalue for any trace-free matrix (or else all eigenvalues could be zero, in which we have the zero matrix).

It follows that every non-zero trace-free Hermitian matrix is necessarily indefinite, and by definition, all your matrices are trace-free and hence indefinite. Therefore a decomposition of the form $A=BB^\dagger$ is never possible for your class of matrices.

Edit: In case the OP didn't mean for the diagonal to be zero the situation is still quite bleak. More general, let $S$ denote the set of all matrices whose non-main diagonals sum to zero.

First, note that $S$ is closed under negation, i.e., $A \in S$ if and only if $-A\in S$. This already tells us there cannot exist a decomposition for every element of $S$, and more assumptions are required.

Another way to see this is that Sylvester's criterion requires that all of the main diagonal elements be non-negative for there to even be a chance for the matrix to be positive semi-definite. So if the main diagonal were unrestricted, any negative element on the diagonal would invalidate the existence of the required decomposition.

The general conclusion is that without further assumptions, no, such a decomposition need not exist.

  • $\begingroup$ I think OP means that the sum of each off diagonal results in 0. Excluding the main-diagonal. $\endgroup$ – P. Siehr Nov 22 '17 at 13:48
  • $\begingroup$ @P.Siehr The OP said each color summed to zero, so I naturally assumed that held for the main diagonal as well. In any case, I've added a few lines in case that's not what he intended. $\endgroup$ – EuYu Nov 22 '17 at 13:59
  • $\begingroup$ Yes, that is why I asked for examples. $\endgroup$ – P. Siehr Nov 22 '17 at 14:36
  • $\begingroup$ @P.Siehr Thank you. You got it right. I mean that the sum of each off diagonal results in 0, and there are no constraints to the main-diagonal. For example, if N=3, A can be represented as$$ \begin{matrix} a & b & 0 \\ b^* & c & -b \\ 0 & -b^* & d \\ \end{matrix} $$ I wonder if Hermitian matrix A with order N≥3 is positive semi-definite (not positive definite, because B is not square). $\endgroup$ – Tony Yan Nov 22 '17 at 17:02
  • $\begingroup$ @EuYu Thank you for your detailed response. First, I mean that the sum of each off diagonal results in 0. Moreover, since A is positive semi-definite, the main diagonal elements are non-negative. So I want to find A under these constraints. Thanks a lot. ^-^ $\endgroup$ – Tony Yan Nov 22 '17 at 17:31

No. The negative of the $1\times 1$ identity matrix, $A = \pmatrix{-1},$ cannot be so written.

  • $\begingroup$ I am sorry for missing the condition that the order of A is more than 2, i.e. N≥3. Thank you. $\endgroup$ – Tony Yan Nov 22 '17 at 13:37

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