Let \(V\) be a finite-dimensional vector space over a subfield of the complex numbers, and let \(f\), \(g\) be skew-symmetric bilinear forms on \(V\). Show that there is an invertible linear operator \(T\) on \(V\) such that \(f(T\alpha, T\beta) = g(\alpha, \beta)\) for all \(\alpha\), \(\beta\) if and only if \(f\) and \(g\) have the same rank.
Solution
We first establish the canonical form needed for the converse.
1. Every skew-symmetric form has a basis in which its matrix is a direct sum of standard \(2\times2\) skew blocks and zeros.
Let \(h\) be a skew-symmetric bilinear form on an \(n\)-dimensional space. If \(h=0\), every basis gives the zero matrix, so there is nothing to prove.
Suppose \(h\ne0\). Choose \(u,v\) with \(h(u,v)\ne0\), and rescale \(v\) so that \[ h(u,v)=1. \] Let \[ H=\operatorname{span}\{u,v\}. \] In the ordered basis \((u,v)\), the restriction of \(h\) to \(H\) has matrix \[ J= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}, \] so it is non-degenerate on \(H\).
Define \[ H^\perp=\{x\in V:h(x,H)=0\}. \] For any \(x\in V\), put \[ h_x=h(x,v)u-h(x,u)v. \] Then \[ h(x-h_x,u)=0, \qquad h(x-h_x,v)=0, \] so \(x-h_x\in H^\perp\). Hence \[ V=H+H^\perp. \] The sum is direct: if \(au+bv\in H^\perp\), pairing with \(u\) and \(v\) gives \(b=0\) and \(a=0\). Thus \[ V=H\oplus H^\perp, \] and the matrix of \(h\) is block diagonal with the block \(J\) and the matrix of \(h|_{H^\perp}\).
Repeat the construction on \(H^\perp\). After finitely many steps we obtain a basis \[ u_1,v_1,\ldots,u_m,v_m,z_1,\ldots,z_s \] for which \[ [h]=J^{\oplus m}\oplus 0_s. \] The remaining restriction is zero, for otherwise another nondegenerate \(2\)-plane could be split off. Consequently \[ \operatorname{rank}h=2m. \]
2. Congruent forms have the same rank.
Assume there is an invertible \(T\) such that \[ f(T\alpha,T\beta)=g(\alpha,\beta) \] for all \(\alpha,\beta\). Choose any basis and let \(P\) be the matrix of \(T\) in that basis. If \(A\) and \(B\) are the matrices of \(f\) and \(g\), respectively, then \[ B=P^tAP. \] Because \(P\) and \(P^t\) are invertible, \[ \operatorname{rank}B=\operatorname{rank}A. \] Thus \[ \operatorname{rank}g=\operatorname{rank}f. \]
3. Skew-symmetric forms of the same rank are congruent.
Assume \[ \operatorname{rank}f=\operatorname{rank}g=2m. \] By step 1, choose an ordered basis \[ \mathfrak B_f=(u_1,v_1,\ldots,u_m,v_m,z_1,\ldots,z_s) \] for which \[ [f]_{\mathfrak B_f}=J^{\oplus m}\oplus0_s, \] and an ordered basis \[ \mathfrak B_g=(u'_1,v'_1,\ldots,u'_m,v'_m,z'_1,\ldots,z'_s) \] for which \[ [g]_{\mathfrak B_g}=J^{\oplus m}\oplus0_s. \] The same value of \(s=n-2m\) occurs because the two forms live on the same space and have equal rank.
Define \(T\) by sending the \(g\)-canonical basis to the \(f\)-canonical basis: \[ T(u'_i)=u_i,\qquad T(v'_i)=v_i,\qquad T(z'_j)=z_j. \] This maps a basis to a basis, so \(T\) is invertible. If \(\alpha,\beta\) have coordinate columns \(x,y\) in \(\mathfrak B_g\), then \(T\alpha,T\beta\) have the same coordinate columns \(x,y\) in \(\mathfrak B_f\). Therefore \[ \begin{aligned} f(T\alpha,T\beta) &=x^t(J^{\oplus m}\oplus0_s)y\\ &=g(\alpha,\beta). \end{aligned} \] Thus the required congruence exists.
Combining steps 2 and 3, \[ \boxed{ \exists\,T\in GL(V): f(T\alpha,T\beta)=g(\alpha,\beta) \iff \operatorname{rank}f=\operatorname{rank}g.} \]