We consider $n\times n$ complex matrices. Let $i_+(A), i_-(A), i_0(A)$ be the number of eigenvalues of $A$ with positive real part, negative real part and pure imaginary. It is well known if two Hermitian matrices $A$ and $B$ are $*-$congruent, then $$(i_+(A), i_-(A), i_0(A))=(i_+(B), i_-(B), i_0(B)).\qquad{(1)}$$ If two general matrices $A$ and $B$ are $*-$congruent, (1) may not hold (can you provide an example?). Moreover, whether a matrix and its transpose are always $*-$congruent?

  • $\begingroup$ What do you mean by "But I have come up a counterexample for this"? Counter-example for what? Also, what do you mean by *-congruent? $\endgroup$ – Yemon Choi Mar 8 '10 at 2:02
  • $\begingroup$ Really sorry. I mean I'd like to see a counterexample such that (1) does not hold when $A$ and $B$ are not Hermitian matrices. $\endgroup$ – Sunni Mar 8 '10 at 2:14
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    $\begingroup$ A and B are $*$-congruent if there is an invertible matrix C such that $B=CAC^*$. $\endgroup$ – Jonas Meyer Mar 8 '10 at 2:45

An answer to the second question: Yes, a square complex matrix is always $*$-congruent to its transpose, according to a more general result proved by Horn and Sergeichuk in "Congruences of a square matrix and its transpose". They prove the result for all fields with involution in characteristic other than 2.

Added: Here's a counterexample for your first question:

$\left( \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$.


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