Let \(k\) be a field, \(V\) a \(k\)-vector space, and \(b\colon V\times V \to k\) a bilinear form. The form \(b\) is nondegenerate if its adjoint map \[ \begin{aligned} V &\to V^{\vee} \\ x &\mapsto b(x, \wait) \end{aligned} \] is injective.
The kernel of the adjoint map is \(\theset{ x \in V \st b(x,y) = 0 \text{ for all } y \in V }\), so \(b\) is nondegenerate if and only if this subspace is zero. If \(\dim_k V = n< \infty\), then \(\dim_k V^{\vee}=n\), so the adjoint map is injective if and only if it is an isomorphism; with respect to a basis \(e_1,\ldots,e_n\) of \(V\) and its dual basis, the adjoint map has matrix the transpose of the Gram matrix \(\bigl(b(e_i,e_j)\bigr)_{i,j}\), so \(b\) is nondegenerate if and only if a Gram matrix of \(b\) is invertible.
See [@Art11].
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D-5BR4DNondegenerate symmetric and Hermitian formsrelated-to