Let \(f\) be a bilinear form on a finite-dimensional vector space \(V\). Suppose \(V_1\) is a subspace of \(V\) with the property that the restriction of \(f\) to \(V_1\) is non-degenerate. Show that \(\operatorname{rank} f \geq \dim V_1\).
Solution
Define the linear map associated with \(f\) by \[ L_f:V\longrightarrow V^*, \qquad L_f(\alpha)(\beta)=f(\alpha,\beta). \] By definition, \[ \operatorname{rank}f=\operatorname{rank}L_f. \]
1. The restriction of \(f\) to \(V_1\) defines an injective map \(V_1\to V_1^*\).
Let \[ L_{f_1}:V_1\longrightarrow V_1^*, \qquad L_{f_1}(\alpha)(\beta)=f(\alpha,\beta) \] for \(\alpha,\beta\in V_1\). If \[ L_{f_1}(\alpha)=0, \] then \[ f(\alpha,\beta)=0 \] for every \(\beta\in V_1\). Since the restriction \(f_1\) is non-degenerate, this forces \(\alpha=0\). Hence \(L_{f_1}\) is injective, and therefore \[ \operatorname{rank}L_{f_1}=\dim V_1. \]
2. The restricted map factors through \(L_f\).
Let \[ i:V_1\hookrightarrow V \] be inclusion and let \[ \rho:V^*\longrightarrow V_1^* \] be restriction of linear functionals. For \(\alpha\in V_1\) and \(\beta\in V_1\), \[ (\rho L_f i)(\alpha)(\beta) =L_f(\alpha)(\beta) =f(\alpha,\beta). \] Thus \[ L_{f_1}=\rho\circ L_f\circ i. \] Consequently \[ \operatorname{rank}L_{f_1} \le \operatorname{rank}L_f. \]
3. Conclude the rank inequality.
Combining steps 1 and 2 gives \[ \dim V_1 =\operatorname{rank}L_{f_1} \le\operatorname{rank}L_f =\operatorname{rank}f. \] Therefore \[ \boxed{\operatorname{rank}f\ge\dim V_1.} \]