Let \(f, g\) be bilinear forms on a finite-dimensional vector space \(V\). Suppose \(g\) is non-singular. Show that there exist unique linear operators \(T_1, T_2\) on \(V\) such that \[ f(\alpha, \beta) = g(T_1\alpha, \beta) = g(\alpha, T_2\beta) \] for all \(\alpha, \beta\).
Solution
For any bilinear form \(h\) on \(V\), define \[ L_h:V\longrightarrow V^*, \qquad L_h(\alpha)(\beta)=h(\alpha,\beta), \] and \[ R_h:V\longrightarrow V^*, \qquad R_h(\beta)(\alpha)=h(\alpha,\beta). \]
1. Nonsingularity of \(g\) makes both \(L_g\) and \(R_g\) isomorphisms.
If \(L_g(\alpha)=0\), then \[ g(\alpha,\beta)=0 \] for every \(\beta\in V\). Nonsingularity of \(g\) forces \(\alpha=0\), so \(L_g\) is injective. Because \(V\) and \(V^*\) have the same finite dimension, \(L_g\) is an isomorphism.
For a bilinear form on a finite-dimensional space, left and right nonsingularity are equivalent. Hence the same argument shows that \(R_g\) is also injective and therefore an isomorphism.
2. Construct \(T_1\) and verify the first factorization.
Define \[ T_1=L_g^{-1}L_f. \] This is a linear operator on \(V\). For every \(\alpha,\beta\in V\), \[ \begin{aligned} g(T_1\alpha,\beta) &=L_g(T_1\alpha)(\beta)\\ &=L_f(\alpha)(\beta)\\ &=f(\alpha,\beta). \end{aligned} \] Thus \[ f(\alpha,\beta)=g(T_1\alpha,\beta). \]
3. Construct \(T_2\) and verify the second factorization.
Define \[ T_2=R_g^{-1}R_f. \] Then for every \(\alpha,\beta\), \[ \begin{aligned} g(\alpha,T_2\beta) &=R_g(T_2\beta)(\alpha)\\ &=R_f(\beta)(\alpha)\\ &=f(\alpha,\beta). \end{aligned} \] Hence \[ f(\alpha,\beta)=g(\alpha,T_2\beta). \]
4. Prove uniqueness of \(T_1\) and \(T_2\).
Suppose \(S_1\) also satisfies \[ f(\alpha,\beta)=g(S_1\alpha,\beta) \] for all \(\alpha,\beta\). Then \[ L_gS_1=L_f=L_gT_1. \] Since \(L_g\) is injective, \[ S_1=T_1. \]
Similarly, if \(S_2\) satisfies \[ f(\alpha,\beta)=g(\alpha,S_2\beta), \] then \[ R_gS_2=R_f=R_gT_2, \] and injectivity of \(R_g\) gives \[ S_2=T_2. \] Therefore the two required operators exist uniquely.