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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.

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  • $\begingroup$ Your reasoning seems correct up until But the entry in the $k$th column of the first row of the RREF matrix is 1. I'm not sure what you're trying to say there, an it doesn't seem correct. $\endgroup$ – Omnomnomnom May 1 '17 at 15:59
  • $\begingroup$ The rank must be even, as is shown in this post: math.stackexchange.com/questions/7978/…. $\endgroup$ – StubbornAtom May 1 '17 at 16:07
  • $\begingroup$ @StubbornAtom that post uses eigenvalues. $\endgroup$ – Omnomnomnom May 1 '17 at 16:08
  • $\begingroup$ @Omnomnomnom Yeah you are right. I got my mistake. But I think I can proceed as follows: Since the $k$th row of $A$ is 0, then its rank is also $0$. This is a contradiction. $\endgroup$ – Adienl May 1 '17 at 16:13
  • $\begingroup$ @Omnomnomnom Yeah you are right. I got my mistake. But I think I can proceed as follows: Since the $k$th row of $A$ is 0, then its rank is also $0$. This is a contradiction. $\endgroup$ – Adienl May 1 '17 at 16:13
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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 $1$.

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  • $\begingroup$ This is cool. Please check my comment. $\endgroup$ – Adienl May 1 '17 at 16:14
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Here's a simple-minded approach. A matrix has rank $1$ iff all its rows are multiples of each other. A skew symmetric matrix looks like $$A=\pmatrix{0&a_{12}&a_{13}&\cdots\\ -a_{12}&0&a_{23}&\cdots\\ -a_{13}&a_{23}&0&\cdots\\ \vdots&\vdots&\vdots&\ddots }.$$ If $a_{12}\ne0$ then row 1 is nonzero, but there is no way that row 2 is a multiple of row 1 since it begins with a nonzero entry. This can be adapted when entry $a_{ij}\ne0$; then row $j$ is no multiple of row $i$.

Of course the rank of a skew-symmetric matrix is even. I don't see a way of adapting this naive approach to prove that.

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  • $\begingroup$ For the generalization: it might help to note that a matrix $A$ has rank $r$ if and only if some $r \times r$ submatrix has a non-zero determinant. $\endgroup$ – Omnomnomnom May 1 '17 at 16:33
  • $\begingroup$ Also, there might be some neat argument in terms of the Pfaffian $\endgroup$ – Omnomnomnom May 1 '17 at 16:37

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