Hurwitz stability status of two matrices

Question

I have a complex symmetric matrix (it is not Hermitian), i.e. $\textbf{A}\in\mathbb{C}^{n\times n}$. Can you prove that $\textbf{A}$ and $\textbf{B}=\textbf{A}+\textbf{A}^*$ have similar Hurwitz stability status, where superscript asterisk denotes complex conjugate? In other words, the number of eigenvalues of $\textbf{A}$ and $\textbf{B}$ with negative real parts are the same. I have tested this conjecture by coding a script in MATLAB and generating random complex symmetric matrices and it seems to be true. However, I cannot prove it. I have searched the net and consulted linear algebra books but they were of no use. Thanks a lot.

Answer 1

Take $$A = \begin{pmatrix} 1 &2\\2&1+4\text i \end{pmatrix} $$ whose only eigenvalue is $1+2\text i$ with multiplicity 2. The matrix $$B = A+A^* = \begin{pmatrix} 2 &4\\4&2 \end{pmatrix} $$ has instead eigenvalues $6$ and $-2$.

Answer 2

This can shown to be true if $B$ is positive/negative definite; otherwise, counterexamples exist.

Let $A=X+iY$ where $X$ and $Y$ are real symmetric. Then $B=2X$. Let $\lambda$ be any eigenvalue of $A$ and $u$ be a corresponding unit eigenvector. Since $X$ and $Y$ are real symmetric, $u^\dagger Xu$ and $u^\dagger Yu$ are real numbers. It follows that $$ \operatorname{Re}(\lambda)=\operatorname{Re}(u^\dagger Au) =\operatorname{Re}(u^\dagger Xu+iu^\dagger Yu) =u^\dagger Xu =\frac12u^\dagger Bu. $$ So, when $B$ is positive (resp. negative) definite, all eigenvalues of $A$ are positive (resp. negative).

This nice agreement in the number of eigenvalues with positive/negative real parts breaks down when $B$ is not definite. The easiest counterexample is given by $$ A=\pmatrix{1\\ i}\pmatrix{1&i}=\pmatrix{1&i\\ i&-1} \quad\text{and}\quad B=2\pmatrix{1\\ &-1}, $$ in which $A$ is nilpotent but $B$ is indefinite.

It does’t help even if you require $A$ to be an invertible diagonalisable matrix and $B$ to be semi-definite. E.g. when $$ A=\pmatrix{0&-\sqrt{2}i\\ -\sqrt{2}i&3} \quad\text{and}\quad B=2\pmatrix{0\\ &3}, $$ $A$ has two positive eigenvalues $1$ and $2$ but $B$ has only one positive eigenvalue.