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?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

