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the group $GU(n,q)$ is usually defined as the group of $n$ by $n$ matrices $X$ over $\mathbb{F}_{q^2}$ such that $X^H X=I_n$, where $I_n$ denotes the identity matrix and $^H$ the hermitian transpose (every entry is replaced by its $q$-th power, and then the transpose is taken)

But what if I were to consider the group of invertible $n$ by $n$ matrices $X$ such that $X^H A X=A$, where $A$ is some $n$ by $n$ matrix (not necessarily hermitian)? How big is this group, and how similar is it to $GU(n,q)$?

Such matrices $X$ will also satisfy $X^H (A+A^H) X=A+A^H$ so I expect the group to be isomorphic to a subgroup of $GU(n,q)$ in many cases?

Thanks in advance

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Two comments: (1) $A+A^H$ is not necessarily similar to $I_n$. Consider $A$ the zero matrix. (2) the set of matrices $X^H AX = A$ is not necessarily a group, unlike the case $A = I_n$. – Willie Wong Feb 1 '13 at 12:05
I agree with you on the first part, and I assume that the problem in part (2) is just that if A is not invertible, X might be singular as well? (I edited the original question) Sorry for being too fast! – fred Feb 1 '13 at 12:16
Maybe I'm missing something trivial, but I force $X$ to be invertible and to satisfy $X^H A X= A$, don't I have a group in that case? Since $X^H A X= A\Longrightarrow A X =X^{-H} A\Longrightarrow A= X^{-H} A X^{-1}=(X^{-1})^H A X^{-1}$. – fred Feb 1 '13 at 12:33
Ah, I was responding to your comment. I see that you edited to required $X$ to be invertible. Yeah, that's fine then. (Removed my comment.) – Willie Wong Feb 1 '13 at 12:45
Exactly, and by Hermitian transpose I mean: applying the involution fixing $\mathbb{F}_{q}$ and then taking the transpose (sorry, I forgot to mention this, and edited my question) – fred Feb 1 '13 at 14:33

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