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I have a couple of questions regarding skew-Hermitian matrices over finite fields. A matrix $A$ over $\mathbb{F}_{q^{2}}$ is skew-Hermitian if $A + A^{*} = 0$, where $A^{*}$ is the conjugate transpose with respect to the involutory field automorphism $x \mapsto x^{q}$.

  1. Is there a restriction on the rank of a skew-Hermitian matrix, as in the case of skew-symmetric matrices? Is this affected by whether the characteristic is $2$, or whether we add the requirement that the diagonal is zero?
  2. If $C$ and $D$ are skew-Hermitian matrices with the same rank, can we say there exists a matrix $A$ such that $A C A^{*} = D$?
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    $\begingroup$ It will certainly depend on whether the polynomial $x^2 - 1$ has a root, or at least how one chooses this "involutory automorphism" $\endgroup$ – Omnomnomnom Oct 22 '15 at 15:37
  • $\begingroup$ Characteristic two does require exceptional handling, because a diagonal matrix with zeros and ones only, viewed as having entries in $\Bbb{F}_4$, is skew-Hermitian according to this definition. And its rank can be anything. $\endgroup$ – Jyrki Lahtonen Oct 22 '15 at 17:33
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    $\begingroup$ Duh. Even in odd characteristic any diagonal matrix with entries from the kernel of the relative trace $x\mapsto x+x^q$ is skew-Hermitian. So all ranks occur without he assumption that the diagonal should be zero. Thinking... $\endgroup$ – Jyrki Lahtonen Oct 22 '15 at 17:39
  • $\begingroup$ @JyrkiLahtonen That's a good way to think about it. I had a nagging worry that they should have even rank as for skew-symmetric/alternating matrices, but I see now that is clearly not the case. $\endgroup$ – Morgan Rodgers Oct 22 '15 at 21:15
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Okay, I think I have managed to solve these questions.

We can have a skew-Hermitian matrix of any rank (thanks Jyrki); we just take an element $\alpha$ such that the trace $T(\alpha) = \alpha+\alpha^{q} = 0$. Then if we want an $n \times n$ skew-Hermitian matrix with rank $k$ we just take $$C = \alpha \begin{bmatrix} I_{k} & 0\\0&0 \end{bmatrix}.$$
If the characteristic of the field is $2$, we can take $\alpha = 1$.

I'm going to actually cop out and ignore my question about what happens when the diagonal is required to be zero. I mentioned this because in dealing with bilinear forms, this is naturally implied with a skew-symmetric matrix in odd characteristic, but needs to be explicitly stated in even characteristic for the bilinear form to define a symplectic geometry. Nothing like this comes up in the skew-symmetric situation.

I will approach the second problem by first considering Hermitian matrices. Let $C$ be an $n \times n$ Hermitian matrix with entries in $\mathbb{F}_{q^{2}}$ having rank $k$ (WLOG assume $k \geq 1$). I want to show that there is an invertible matrix $A$ such that $$ ACA^{*} = \begin{bmatrix} I_{k} & 0\\0 & 0 \end{bmatrix}$$ (block matrix representation where the bottom right block is $(n-k)\times (n-k)$). The key here (and the motivation for the question) is to think of $C$ as representing a Hermitian bilinear form $B(u,v) = u C v^{*}$, then the matrix $A C A^{*}$ represents this bilinear form with respect to a different basis (basis vectors being the rows of $A$). The collection of vectors $u$ such that $u C v^{*} = 0$ for all $v \in V = \mathbb{F}_{q^{2}}^{n}$ is a a subspace called the radical, so call this $W$. It is clear that $W = \mathrm{null}(C)$ so $\dim(W) = n-k$; therefore we will say that $\{w_{1}, \ldots, w_{n-k}\}$ is a basis for $W$. Then $V = V_{0} \oplus W$ for some subspace $V_{0} \leq V$ for which the Hermitian form restricted to $V_{0}$ is nondegenerate (and $\dim(V_{0}) = k$).

Now our bilinear form is nondegenerate on $V_{0}$, so we can find a vector $v \in V_{0}$ with $vCv^{*} = c \neq 0$. It can also be seen that $c = c^{q}$ (because $(vCv^{*})^{*} = vCv^{*}$), therefore $c \in \mathbb{F}_{q}$. The norm map $N:\mathbb{F}_{q^{2}} \to \mathbb{F}_{q}$ is surjective, so there exists an $\omega \in \mathbb{F}_{q^{2}}$ with $N(\omega) = \omega^{q+1} = c$, and so putting $v_{1} = \frac{1}{\omega}v$ we have $v_{1}Cv_{1}^{*} = 1$, and $V_{0} = \langle v \rangle \oplus \langle v \rangle^{\perp}$. Using induction on $\dim(V_{0})=k$ lets us find a basis $\{v_{1}, \ldots, v_{k}\}$ with $v_{i}Cv_{i}^{*} = 1$ for all $i$, and $v_{i} C v_{j}^{*} = 0$ for all $i \neq j$. Now finally we can put $$A = \begin{bmatrix} \phantom{\ldots \ldots} v_{1} \phantom{\ldots \ldots} \\ \vdots \\ v_{k} \\ w_{1} \\ \vdots \\ w_{n-k} \end{bmatrix}$$ and we have $$ACA^{*} = \begin{bmatrix} I_{k} & 0 \\ 0 & 0 \end{bmatrix}.$$

We will now prove the analogous result for skew-Hermitian matrices. Take $\alpha \in \mathbb{F}_{q^{2}}$ such that $T(\alpha) = \alpha^{q}+\alpha = 0$. WLOG we can assume that $N(\alpha) = \alpha^{q+1} = 1$, because otherwise if $N(\alpha) = c$ then we can replace $\alpha$ with $\frac{\alpha}{\sqrt{c}}$. Let $C$ be skew-Hermitian so $C^{*} = -C$. We want to show that there exists a matrix $A$ such that $$ACA^{*} = \alpha \begin{bmatrix} I_{k} & 0\\0&0\end{bmatrix},$$ where $k$ is the rank of $C$. Now $(\alpha^{q} C)^{*} = \alpha C^{*} = \alpha^{q} C$, that is, $\alpha^{q} C$ is Hermitian, so there exists a matrix $A$ such that $$A (\alpha^{q} C) A^{*} = \alpha^{q} A C A^{*} = \begin{bmatrix} I_{k} & 0 \\ 0 & 0 \end{bmatrix}.$$ Multiplying both sides by $\alpha$ we have $$\alpha^{q+1} A C A^{*} = A C A^{*} = \alpha \begin{bmatrix} I_{k} & 0 \\ 0 & 0 \end{bmatrix}.$$

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