Remarks (See here:
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.
$\blacksquare$
where
$$ A_i = \alpha_i \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) $$with $\alpha_i$ real numbers, $i = 1,\ldots,t$.
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Author of the notes: Antonio J. Pan-Collantes
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