Call the form \(f\) (left) non-degenerate if 0 is the only vector \(\alpha\) such that \(f(\alpha, \beta) = 0\) for all \(\beta\). Let \(f\) be a form on an inner product space \(V\). Prove that \(f\) is non-degenerate if and only if the associated linear operator \(T_f\) (Theorem 1) is non-singular.
Solution
By definition of the associated operator, \[ f(\alpha,\beta)=(T_f\alpha\mid\beta) \] for all \(\alpha,\beta\in V\).
1. The left radical of \(f\) is exactly \(\ker T_f\).
Suppose first that \(\alpha\in\ker T_f\). Then \(T_f\alpha=0\), so \[ f(\alpha,\beta) =(T_f\alpha\mid\beta) =0 \] for every \(\beta\).
Conversely, suppose \[ f(\alpha,\beta)=0 \] for every \(\beta\). Then \[ (T_f\alpha\mid\beta)=0 \] for every \(\beta\). Taking \(\beta=T_f\alpha\) gives \[ \|T_f\alpha\|^2=0, \] hence \(T_f\alpha=0\). Thus \[ \{\alpha:f(\alpha,\beta)=0\text{ for all }\beta\} =\ker T_f. \]
2. Non-degeneracy is equivalent to nonsingularity of \(T_f\).
The form \(f\) is left non-degenerate exactly when its left radical is \(\{0\}\). By step 1 this is equivalent to \[ \ker T_f=\{0\}, \] so \(T_f\) is injective. Since \(V\) is finite-dimensional, an injective linear operator \(V\to V\) is invertible. Therefore \[ \boxed{f\text{ is non-degenerate}\iff T_f\text{ is nonsingular}.} \]