Let \(A\) be a real skew-symmetric matrix, so \(A_{ij}=-A_{ji}\). Prove that \(A\) has even rank.
Solution
Let \(A\) be an \(n\times n\) matrix and let \(V=\RR^n\).
1. The bilinear form\[ \omega:V\times V\longrightarrow\RR, \qquad \omega(u,v)=u^TAv, \]is alternating.
Since \(A^T=-A\), \[ \omega(v,u) = v^TAu = u^TA^Tv = -u^TAv = -\omega(u,v). \] Taking \(u=v\) gives \[ \omega(v,v)=-\omega(v,v), \] so \(\omega(v,v)=0\) over \(\RR\).
2. The radical of \(\omega\) is exactly \(\ker A\).
For \(u\in V\), \[ \omega(u,v) = u^TAv = -(Au)^Tv. \] Thus \(\omega(u,v)=0\) for every \(v\in V\) if and only if \(Au=0\).
3. The form \(\omega\) induces a nondegenerate alternating form\(\overline\omega\) on\[ \overline V \coloneqq V/\ker A, \]and\[ \dim\overline V=\operatorname{rank}A. \]
Because \(\ker A\) is the radical by step 2, changing either representative by an element of \(\ker A\) does not change the value of \(\omega\). Hence \(\omega\) descends to the quotient. The induced form has zero radical by construction, so it is nondegenerate. Finally, rank-nullity gives \[ \dim(V/\ker A) = \dim V-\dim\ker A = \operatorname{rank}A. \]
4. Every finite-dimensional real vector space carrying anondegenerate alternating bilinear form has even dimension.
We argue by induction on the dimension. Dimension \(0\) is even.
Let \(W\) be nonzero and let \(\beta\) be a nondegenerate alternating form on \(W\). Choose \(0\neq v\in W\). By nondegeneracy, there exists \(w\in W\) such that \[ \beta(v,w)\neq0. \] Set \[ U=\operatorname{span}\{v,w\}. \] In the basis \((v,w)\), the matrix of \(\beta|_U\) is \[ \begin{pmatrix} 0&c\\ -c&0 \end{pmatrix}, \qquad c=\beta(v,w)\neq0, \] so \(\beta|_U\) is nondegenerate.
Let \[ U^\perp \coloneqq \{x\in W:\beta(x,u)=0\text{ for every }u\in U\}. \] Because \(\beta|_U\) is nondegenerate, every \(x\in W\) has a unique decomposition \[ x=u+y, \qquad u\in U, \quad y\in U^\perp. \] Indeed, the map \[ U\longrightarrow U^*, \qquad u\longmapsto\beta(u,-)|_U, \] is an isomorphism. Hence for each \(x\in W\) there is a unique \(u\in U\) such that \[ \beta(u,-)|_U=\beta(x,-)|_U. \] Then \(y=x-u\) lies in \(U^\perp\). Uniqueness also shows \(U\cap U^\perp=\{0\}\). Thus \[ W=U\oplus U^\perp. \] Moreover, the restriction of \(\beta\) to \(U^\perp\) is nondegenerate: if \(y\in U^\perp\) is orthogonal to all of \(U^\perp\), then it is orthogonal to both summands of \(W\), hence to all of \(W\), so \(y=0\).
Therefore \[ \dim W = 2+\dim U^\perp. \] By induction, \(\dim U^\perp\) is even, so \(\dim W\) is even.
5. The rank of \(A\) is even.
By step 3, \(\overline V\) carries a nondegenerate alternating form and \[ \dim\overline V=\operatorname{rank}A. \] Step 4 says that \(\dim\overline V\) is even.
6. Q.E.D.
Step 5 is the required conclusion.