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?
 A: Notice that over a field of characteristic 2, the matrix
$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$$
is skew-symmetric, and has rank $3$. So I think you are really looking for a proof that an alternating matrix (skew-symmetric with zeros on the diagonal) has even rank.
The proof for this usually comes up in looking at the theory of bilinear forms, there are a few references mentioned in the comments that contain proofs, but another option is in "Classical Groups and Geometric Algebra" by Larry Grove.
A: Here's an elementary proof.
Suppose that $A$ is a skew-symmetric matrix of rank $r$ and dimension $n \times n$. Now $r$ could very well be zero, and since zero is an even number, then $A$ has an even rank. So assume instead that $r > 0$. Consequently, we can pick out exactly $r$ rows, say those with the indices $i_1, i_2, ..., i_r$, which span the entire row space. Given that for a skew-symmetric matrix each column is equal to $-1$ times the transpose of the corresponding row, therefore every column of the matrix can be expressed as a linear combination of the columns with indices $i_1, ..., i_r$ in the exact same way that the corresponding row is expressed as a linear combination of the rows with these same indices. We know that if we remove a row/column of a matrix that is in the span of the remaining rows/columns, the rank does not change. Thus, we can remove all the $n - r$ rows and $n - r$ columns remaining and not change the rank. Due to symmetry, every time we remove a row, we remove its corresponding column. This way, we have preserved the structure of the matrix. The resultant submatrix $A'$ has dimensions $r \times r$ and full rank $r$.
Suppose for the sake of contradiction that $r$ is odd. We know that $A'^T = -A'$. Then the determinant of this submatrix is
$$\det(A') = \det(A'^T) = \det(-A') = (-1)^{r} \cdot \det(-A') = -\det(A')$$
and therefore $\det(A') = 0$. This contradicts our assumption that $r \neq 0$. Hence, $r$ is an even natural number.
A: You can see Hoffman's book on linear algebra last chapter on Bilinear forms which says rank of skew symmetric matrix is always even.
A: $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
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.
