Let \(V\) be a finite-dimensional vector space over a subfield \(F\) of the complex numbers, and let \(S\) be the set of all symmetric bilinear forms on \(V\).
Show that \(S\) is a subspace of \(L(V, V, F)\).
Find \(\dim S\).
Let \(Q\) be the set of all quadratic forms on \(V\).
Show that \(Q\) is a subspace of the space of all functions from \(V\) into \(F\).
Describe explicitly an isomorphism \(T\) of \(Q\) onto \(S\), without reference to a basis.
Let \(U\) be a linear operator on \(V\) and \(q\) an element of \(Q\). Show that the equation \((U^{\dagger}q)(\alpha) = q(U\alpha)\) defines a quadratic form \(U^{\dagger}q\) on \(V\).
If \(U\) is a linear operator on \(V\), show that the function \(U^{\dagger}\) defined in part (e) is a linear operator on \(Q\). Show that \(U^{\dagger}\) is invertible if and only if \(U\) is invertible.
Solution
Let \[ n=\dim V. \] Since \(F\) is a subfield of \(\mathbb C\), its characteristic is zero and in particular \(2\) is invertible in \(F\).
1. The symmetric bilinear forms form a vector subspace.
If \(f,g\in S\) and \(a,b\in F\), then \(af+bg\) is bilinear. Moreover \[ \begin{aligned} (af+bg)(\alpha,\beta) &=a f(\alpha,\beta)+b g(\alpha,\beta)\\ &=a f(\beta,\alpha)+b g(\beta,\alpha)\\ &=(af+bg)(\beta,\alpha). \end{aligned} \] Thus \(af+bg\in S\). The zero bilinear form is symmetric, so \(S\) is a subspace of \(L(V,V,F)\).
2. Compute \(\dim S\).
Choose a basis \(e_1,\ldots,e_n\) of \(V\). A symmetric bilinear form is represented in this basis by a symmetric \(n\times n\) matrix. Such a matrix is determined freely by the \(n\) diagonal entries and the \[ \binom n2 \] entries above the diagonal. Hence \[ \boxed{\dim S=n+\binom n2=\frac{n(n+1)}2.} \]
3. The quadratic forms form a vector subspace of all functions \(V\to F\).
Every quadratic form is of the form \[ q_f(\alpha)=f(\alpha,\alpha) \] for a symmetric bilinear form \(f\). If \(q_f,q_g\in Q\) and \(a,b\in F\), then \[ \begin{aligned} (a q_f+b q_g)(\alpha) &=a f(\alpha,\alpha)+b g(\alpha,\alpha)\\ &=(af+bg)(\alpha,\alpha). \end{aligned} \] Since \(af+bg\) is symmetric by step 1, \(a q_f+b q_g\in Q\). The zero function is the quadratic form associated with the zero bilinear form. Thus \(Q\) is a vector subspace of the function space \(F^V\).
4. Construct the basis-free isomorphism \(T:Q\to S\).
For \(q\in Q\), define \[ (Tq)(\alpha,\beta) =\frac12\bigl(q(\alpha+\beta)-q(\alpha)-q(\beta)\bigr). \] This formula makes no reference to a basis.
If \(q=q_f\) for a symmetric bilinear form \(f\), then \[ \begin{aligned} q(\alpha+\beta)-q(\alpha)-q(\beta) &=f(\alpha+\beta,\alpha+\beta) -f(\alpha,\alpha)-f(\beta,\beta)\\ &=f(\alpha,\beta)+f(\beta,\alpha)\\ &=2f(\alpha,\beta). \end{aligned} \] Hence \[ Tq_f=f. \] In particular \(Tq\) is always a symmetric bilinear form.
The map \(T\) is linear because the displayed polarization formula is linear in \(q\). Its inverse is \[ S\longrightarrow Q, \qquad f\longmapsto q_f, \qquad q_f(\alpha)=f(\alpha,\alpha), \] because \(Tq_f=f\). Therefore \[ \boxed{T:Q\overset\sim\longrightarrow S} \] is the required isomorphism.
5. Pullback by \(U\) preserves quadratic forms.
Let \(q=q_f\) with \(f\in S\). Define \[ h(\alpha,\beta)=f(U\alpha,U\beta). \] Because \(U\) is linear and \(f\) is symmetric bilinear, \(h\) is symmetric bilinear. Its associated quadratic form is \[ q_h(\alpha) =h(\alpha,\alpha) =f(U\alpha,U\alpha) =q(U\alpha) =(U^\dagger q)(\alpha). \] Thus \(U^\dagger q\in Q\).
6. The map \(U^\dagger:Q\to Q\) is linear.
For \(q_1,q_2\in Q\), scalars \(a,b\), and \(\alpha\in V\), \[ \begin{aligned} U^\dagger(aq_1+bq_2)(\alpha) &=(aq_1+bq_2)(U\alpha)\\ &=a q_1(U\alpha)+b q_2(U\alpha)\\ &=(aU^\dagger q_1+bU^\dagger q_2)(\alpha). \end{aligned} \] Hence \(U^\dagger\) is linear.
7. If \(U\) is invertible, then \(U^\dagger\) is invertible.
If \(U\) is invertible, then for every \(q\in Q\) and \(\alpha\in V\), \[ \begin{aligned} ((U^{-1})^\dagger U^\dagger q)(\alpha) &=(U^\dagger q)(U^{-1}\alpha)\\ &=q(UU^{-1}\alpha)\\ &=q(\alpha). \end{aligned} \] Similarly \[ U^\dagger (U^{-1})^\dagger q=q. \] Thus \[ (U^\dagger)^{-1}=(U^{-1})^\dagger. \]
8. If \(U\) is singular, then \(U^\dagger\) is singular.
Suppose \(U\) is not invertible. Since \(V\) is finite-dimensional, \(\operatorname{im}U\) is a proper subspace. Choose a nonzero functional \(L\in V^*\) that vanishes on \(\operatorname{im}U\). Then \[ q(\alpha)=L(\alpha)^2 \] is a nonzero quadratic form: it is associated with the symmetric bilinear form \[ f(\alpha,\beta)=L(\alpha)L(\beta). \] But for every \(\alpha\), \[ (U^\dagger q)(\alpha) =q(U\alpha) =L(U\alpha)^2 =0. \] Thus \(U^\dagger q=0\) with \(q\ne0\), so \(U^\dagger\) is not injective and hence not invertible.
Combining steps 7 and 8 gives \[ \boxed{U^\dagger\text{ is invertible}\iff U\text{ is invertible}.} \]