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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?

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4… This looks like what you want. (This took ten seconds' googling...!) – Billy Aug 15 '11 at 22:44
Another source is Chapter XV, Section 8 of Lang's Algebra. – Dylan Moreland Aug 15 '11 at 22:48

This is not an answer, but a remark. The rank of a skew symmetric matrix is not always even. It really depends on the ground field. Over a finite field of characteristic 2 (i.e. $GF(2)$; see here for a brief description), we have $1=-1$. Hence the matrix $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}=-\begin{pmatrix}1&1\\1&1\end{pmatrix}$ is skew symmetric. Yet $A=\begin{pmatrix}1\\ 1\end{pmatrix}\begin{pmatrix}1&1\end{pmatrix}$ is also of rank 1.

That said, a complex skew symmetric matrix does have an even rank, as proved in the monographs mentioned by Billy and Dylan Moreland in the comments to your question.

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Though this is old, I would add that the notion of an alternating matrix is usually used in place of skew-symmetric in characteristic 2 (usually defined in terms of the associated bilinear form). This adds the requirement that the matrix has a $0$ diagonal. In this setting, the matrix will indeed have even rank. – Morgan Rodgers Oct 21 '15 at 22:53

You can see Hoffman's book on linear algebra last chapter on Bilinear forms which says rank of skew symmetric matrix is always even.

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An alternate proof to this :

Skew Symmetric matrix follows $A = -A^T$.

Now, Suppose there exists at least one non zero quantity "a" such that a largest non singular sub matrix exists :

\begin{pmatrix} 0 & a \\ -a & 0 \\ \end{pmatrix} The order of the largest non singular sub matrix in a given matrix is defined as the rank of the matrix . Clearly it's rank = 2. Now, you can generalise for presence of more non zero entities in the matrix.

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