I thought of this argument to show that the rank of a skew-symmetric matrix is even. I'd appreciate it if someone could look over it and let me know whether or not it looks correct.

Let $A$ be an $n$x$n$ skew-symmetric matrix over the field $\mathbb{F}$ (where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$). First, assume $\text{det}(A)\not=0$. then $\beta(x,y)=y^TAx$ is a non-degenerate, skew-symmetric bilinear form on $\mathbb{F}^n$. Since the symplectic space $(\mathbb{F}^n,\beta)$ has even dimension, $n$ is even $\implies$ $\text{rank}(A)=n$ is even.

Now suppose $\text{det}(A)=0$. $\text{rank}(A)=m<n$ and so $A$ has $m$ linearly independent columns. Let $A_{i*}$ be the $i$th row of $A$ and $A_{*,i}$ be the $i$th column of $A$. Since $A$ is skew-symmetric, $C=\{A_{*i_1}\,...,A_{*i_m}\}$ is linearly independent $\iff$ $B=\{A_{i_1*},..., A_{i_m*}\}$. Hence, the submatrix $A_{BC}$ obtained by using the rows of $B$ and the columns of $C$ is $m$ by $m$ and has rank $m$. Also, $A_{BC}$ is skew symmetric, so case $1$ implies $m$ is even. Hence, $\text{rank}(A)=\text{rank}(A_{BC})$ is even.

  • $\begingroup$ I'm not really familiar with the type of arguments you're using (so I can't really valid your proof). Nevertheless the proof in this answer looks quite simple although the argument seem a bit different. $\endgroup$ – Surb Jul 24 '17 at 23:01
  • $\begingroup$ IMO more detail should be given showing that $A_{BC}$ has rank $m$. The fact that rows $i_1,...,i_m$ of $A$ are linearly independent doesn't always imply that rows $i_1,...,i_m$ of $A_{*C}$ are (eg all rows of $I_n$ are linearly independent, but they don't remain so if you drop a column). $\endgroup$ – stewbasic Jul 24 '17 at 23:11
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    $\begingroup$ "Since the symplectic space has even dimension" -- In my opinion invoking this is circular, as the proof of this is essentially the statement in question. $\endgroup$ – Christopher A. Wong Jul 24 '17 at 23:24

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