I was thinking of using the RREF of $A$, where $A$ is an $n \times n$ skew symmetric matrix.
So, suppose that it's rank is 1. then the first row has at least one non zero element(which is 1), while all other rows are 0. Now we can get back to original the skew symmetric matrix by applying the required row operations. Now the rows of the original matrix must be a multiple of the 1st row of the RREF of A. But the entry in the $k$th column of the first row of the RREF matrix is 1. Hence this implies that the $k$th row is $\theta \in \mathbb{R^n}$. So, $a_{k1}=0$. Hence $a_{1k}=0$. This is a contradiction. Hence the rank cannot be 1.
Is this correct? Please do not use eigenvalues if you post an answer.
Thank you for reading.
Best Answer
I don't think your proof is correct, as I point out in a comment. Here's a nice approach:
If $A$ is rank $1$, then $A = uv^T$ for non-zero column vectors $u,v$ with $n$ entries. If $A$ is skew-symmetric, we must have $$ A^T = -A \implies \\ (uv^T)^T = -uv^T\\ vu^T = -uv^T $$ The column space of these matrices is the same. The column space of $vu^T$ is the span of $v$, whereas the column space of $uv^T$ is the span of $u$. So, we must have $v = ku$ for some $k \in \Bbb R$. So, the last equation becomes $$ kuu^T = -kuu^T $$ since $u \neq 0$, we conclude that $k = 0$, which means that $v = 0$, which means that $A = 0$. But this contradicts our assumption that $A$ has rank $1$.