The rank of skew-symmetric matrix is even I know that the rank of a skew-symmetric matrix is even.  I just need to find a published proof for it.  Could anyone direct me to a source that could help me?

$\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
If $K$ is a real skew symmetric matrix, typically it is proven by showing $K$ has even non-zero eigenvalues. Below is a proof using the Hermitian form of $K$.
Since $\rank(K) = r$, there exist order $n$ invertible matrices $P$ and $Q$, such that $K = P\diag(I_{(r)}, 0)Q$. By $K = -K'$, $Q'\diag(-I_{(r)}, 0)P' = P\diag(I_{(r)}, 0)Q$, i.e.,
\begin{align*}
P^{-1}Q\diag(-I_{(r)}, 0) = \diag(I_{(r)}, 0)(P^{-1}Q)'. 
\end{align*}
Let $P^{-1}Q' = R = \begin{pmatrix} R_{11} & R_{12} \\ R_{21} & R_{22} \end{pmatrix}$, where $R_{11}$ is an order $r$ matrix. Substituting it to the above equation yields
$R_{11}' = -R_{11}, R_{21} = 0$. Therefore,
\begin{align*}
P^{-1}Q' = \begin{pmatrix} R_{11} & R_{12} \\ 0 & R_{22} \end{pmatrix}, \text{ i.e., } Q' = P\begin{pmatrix} R_{11} & R_{12} \\ 0 & R_{22} \end{pmatrix},
\end{align*}
whence $Q = \begin{pmatrix} R_{11}' & 0 \\ R_{12}' & R_{22}' \end{pmatrix}$. Consequently,
\begin{align*}
K = P\begin{pmatrix} I_{(r)} & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} R_{11}' & 0 \\ R_{12}' & R_{22}' \end{pmatrix}P'
= P\begin{pmatrix} R_{11}' & 0 \\ 0 & 0 \end{pmatrix}P'.
\end{align*}
Since $P$ is invertible, $\rank(K) = \rank(\diag(R_{11}, 0)) = \rank(R_{11}) = r$, whence $R_{11}$ is an order $r$ invertible matrix. Taking
determinants on both sides of $R_{11}' = -R_{11}$ yields $\det(R_{11}') = (-1)^r\det(R_{11})$, hence $r$ must be an even number, otherwise $\det(R_{11}) = 0$, which contradicts with $R_{11}$ is invertible.
