Show that the result of Exercise 12 is valid for symmetric bilinear forms on a complex vector space, but is not valid for symmetric bilinear forms on a real vector space.
Solution
1. Over \(\mathbb C\), every symmetric bilinear form is congruent to\(\operatorname{diag}(I_r,0)\), where \(r\) is its rank.
Let \(f\) be a symmetric bilinear form on a finite-dimensional complex vector space \(V\).
If \(f=0\), the assertion is immediate. Otherwise there is a vector \(u\) with \(f(u,u)\ne0\). Indeed, if \(f(v,v)=0\) for every \(v\), then symmetry and polarization give \[ 2f(x,y)=f(x+y,x+y)-f(x,x)-f(y,y)=0 \] for all \(x,y\), forcing \(f=0\).
Choose such a vector \(u\). Its one-dimensional span is non-degenerate, and \[ V=\mathbb C u\oplus u^\perp, \qquad u^\perp=\{x:f(x,u)=0\}. \] To see the decomposition explicitly, for \(x\in V\) put \[ a=\frac{f(x,u)}{f(u,u)}. \] Then \(x-au\in u^\perp\), while the intersection \(\mathbb C u\cap u^\perp\) is zero. Repeating this construction on the orthogonal complement yields a basis in which \[ [f]=\operatorname{diag}(c_1,\ldots,c_r,0,\ldots,0), \] where each \(c_i\ne0\) and \(r=\operatorname{rank}f\).
Because the field is \(\mathbb C\), choose \(s_i\in\mathbb C\) with \[ s_i^2=c_i. \] Rescaling the corresponding basis vector by \(1/s_i\) changes the \(i\)th diagonal entry from \(c_i\) to \(1\). Hence there is a basis in which \[ \boxed{[f]=\operatorname{diag}(I_r,0).} \]
2. Complex symmetric forms are congruent if and only if they have the same rank.
Congruence by an invertible operator preserves rank because, in matrices, it has the form \[ B=P^tAP \] with \(P\) invertible.
Conversely, let \(f\) and \(g\) be complex symmetric forms of the same rank \(r\). By step 1, choose bases \(\mathfrak B_f\) and \(\mathfrak B_g\) in which both forms have matrix \[ \operatorname{diag}(I_r,0). \] Define \(T\) to map the \(g\)-basis to the \(f\)-basis. Then \(T\) is invertible and, for all \(\alpha,\beta\), \[ f(T\alpha,T\beta)=g(\alpha,\beta). \] Thus the analogue of Exercise 12 is valid over \(\mathbb C\).
3. Rank does not classify real symmetric bilinear forms up to congruence.
On the one-dimensional real vector space \(\mathbb R\), consider \[ f(x,y)=xy, \qquad g(x,y)=-xy. \] Both forms have rank \(1\).
Every invertible real linear operator has the form \[ T(x)=tx \] for some \(t\in\mathbb R\setminus\{0\}\). Then \[ f(Tx,Ty)=t^2xy. \] If this were equal to \(g(x,y)=-xy\) for all \(x,y\), we would have \[ t^2=-1, \] which has no real solution. Hence \(f\) and \(g\) have the same rank but are not congruent over \(\mathbb R\).
Therefore the rank-only classification fails over the reals; the signs of the nonzero diagonal terms carry additional information.