Let \(k\) be a field, let \(V\) be a finite-dimensional \(k\)-vector space with ordered basis \(\mathcal B = (e_1, \ldots, e_n)\), and let \(b\colon V\times V\to k\) be a bilinear form. The Gram matrix of \(b\) with respect to \(\mathcal B\) is \[ G \coloneqq \qty{ b(e_i, e_j) }_{1\leq i,j\leq n}. \]
Let \(k\), \(V\), \(\mathcal B\), \(b\), and \(G\) be as in the definition, and write \([x]_{\mathcal B}\in k^n\) for the coordinate vector of \(x\in V\).
For all \(x, y\in V\), \(b(x, y) = [x]_{\mathcal B}^t\, G\, [y]_{\mathcal B}\).
If \(\mathcal B'\) is another ordered basis and \(P\in\GL_n(k)\) satisfies \([x]_{\mathcal B} = P[x]_{\mathcal B'}\) for all \(x\in V\), then the Gram matrix of \(b\) with respect to \(\mathcal B'\) is \(P^t G P\). In particular, \(\det G\) is well defined up to multiplication by the square of a nonzero element of \(k\).
\(b\) is nondegenerate if and only if \(G\) is invertible.
See [@Art11].
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