Let \(V\) be an \(n\)-dimensional vector space over a subfield \(F\) of \(C\). Prove the following.
The equation \((Pf)(\alpha, \beta) = \frac{1}{2}f(\alpha, \beta) - \frac{1}{2}f(\beta, \alpha)\) defines a linear operator \(P\) on \(L(V, V, F)\).
\(P^2 = P\), i.e., \(P\) is a projection.
\(\operatorname{rank} P = \frac{n(n-1)}{2}\); \(\operatorname{nullity} P = \frac{n(n+1)}{2}\).
If \(U\) is a linear operator on \(V\), the equation \((U^{\dagger}f)(\alpha, \beta) = f(U\alpha, U\beta)\) defines a linear operator \(U^{\dagger}\) on \(L(V, V, F)\).
For every linear operator \(U\), the projection \(P\) commutes with \(U^{\dagger}\).
Solution
Write \(\mathcal B=L(V,V,F)\) for the vector space of bilinear forms on \(V\).
1. The formula in part (a) defines a linear operator \(P:\mathcal B\to\mathcal B\).
For \(f\in\mathcal B\), define \[ (Pf)(\alpha,\beta) =\frac12\bigl(f(\alpha,\beta)-f(\beta,\alpha)\bigr). \] Because \(f\) is bilinear, this expression is bilinear in \((\alpha,\beta)\), so \(Pf\in\mathcal B\). If \(c,d\in F\) and \(f,g\in\mathcal B\), then \[ P(cf+dg)=cPf+dPg \] by direct expansion. Hence \(P\) is linear.
2. The operator \(P\) is a projection.
For every \(f\), \[ (Pf)(\beta,\alpha)=-(Pf)(\alpha,\beta), \] so \(Pf\) is skew-symmetric. Therefore \[ \begin{aligned} (P^2f)(\alpha,\beta) &=\frac12\bigl((Pf)(\alpha,\beta)-(Pf)(\beta,\alpha)\bigr)\\ &=\frac12\bigl((Pf)(\alpha,\beta)+(Pf)(\alpha,\beta)\bigr)\\ &=(Pf)(\alpha,\beta). \end{aligned} \] Thus \(P^2=P\).
3. The image of \(P\) is the space of skew-symmetric forms and the kernel is the space of symmetric forms.
Step 2 already shows that every \(Pf\) is skew-symmetric. Conversely, if \(g\) is skew-symmetric, then \[ (Pg)(\alpha,\beta) =\frac12(g(\alpha,\beta)-g(\beta,\alpha)) =g(\alpha,\beta), \] so \(g\in\operatorname{im}P\). Hence \[ \operatorname{im}P=\{\text{skew-symmetric forms}\}. \]
Also, \[ Pf=0 \] if and only if \[ f(\alpha,\beta)=f(\beta,\alpha) \] for all \(\alpha,\beta\), i.e. exactly when \(f\) is symmetric. Thus \[ \ker P=\{\text{symmetric forms}\}. \]
4. Compute the rank and nullity of \(P\).
Choose a basis \(e_1,\ldots,e_n\) of \(V\). A bilinear form is determined by its matrix \((f(e_i,e_j))\), so \(\mathcal B\) has dimension \(n^2\).
A skew-symmetric form has zero diagonal entries and is determined freely by the entries with \(i<j\); the entries with \(j<i\) are their negatives. Hence the space of skew-symmetric forms has dimension \[ \binom n2=\frac{n(n-1)}2. \] By step 3, \[ \boxed{\operatorname{rank}P=\frac{n(n-1)}2}. \]
Likewise, a symmetric form is determined freely by the diagonal entries and the entries above the diagonal, giving \[ n+\binom n2=\frac{n(n+1)}2 \] parameters. Thus \[ \boxed{\operatorname{nullity}P=\frac{n(n+1)}2}. \]
5. Pullback by \(U\) defines a linear operator \(U^\dagger\) on \(\mathcal B\).
For \(f\in\mathcal B\), set \[ (U^\dagger f)(\alpha,\beta)=f(U\alpha,U\beta). \] Since \(U\) is linear and \(f\) is bilinear, \(U^\dagger f\) is bilinear. Moreover for scalars \(c,d\) and forms \(f,g\), \[ U^\dagger(cf+dg)=cU^\dagger f+dU^\dagger g. \] Hence \(U^\dagger\) is a linear operator on \(\mathcal B\).
6. The operators \(P\) and \(U^\dagger\) commute.
For every \(f\in\mathcal B\) and \(\alpha,\beta\in V\), \[ \begin{aligned} (PU^\dagger f)(\alpha,\beta) &=\frac12\bigl(f(U\alpha,U\beta)-f(U\beta,U\alpha)\bigr)\\ &=(Pf)(U\alpha,U\beta)\\ &=(U^\dagger Pf)(\alpha,\beta). \end{aligned} \] Therefore \[ \boxed{PU^\dagger=U^\dagger P}. \]