Let \(k\) be a field, \(V\) a finite-dimensional \(k\)-vector space, \(\inner{\wait}{\wait}\) a bilinear form on \(V\), and \(\mathbf B = (v_1, \ldots, v_n)\) an ordered basis of \(V\). The Gram matrix of the form with respect to \(\mathbf B\), also called the matrix of the form, is \[ A = (a_{ij}), \qquad a_{ij} \coloneqq \inner{v_i}{v_j}. \]
If \(X,Y\in k^n\) are the coordinate vectors of \(v,w\in V\) with respect to \(\mathbf B\), then \[ \inner{v}{w} = X^t A Y. \] Consequently, for a fixed basis \(\mathbf B\), the assignment of the Gram matrix is a bijection from bilinear forms on \(V\) to \(\Mat_{n\times n}(k)\), and a form is symmetric if and only if its Gram matrix is symmetric.
Writing \(v=\sum_i x_iv_i\) and \(w=\sum_j y_jv_j\), bilinearity gives \(\inner{v}{w}=\sum_{i,j}x_iy_j\inner{v_i}{v_j}=X^tAY\). Conversely, every \(A\in\Mat_{n\times n}(k)\) defines the bilinear form \((v,w)\mapsto X^tAY\), whose Gram matrix is \(A\). The form is symmetric if and only if \(a_{ij}=a_{ji}\) for all \(i,j\), by the displayed formula.
If \(\mathbf B'\) is another ordered basis and \(P\in\GL_n(k)\) is the change-of-basis matrix, so that coordinates satisfy \(X=PX'\), then the Gram matrix with respect to \(\mathbf B'\) is \(P^t A P\). This preserves the rank of \(A\) and, for \(k=\RR\) and a symmetric form, its signature, but not its eigenvalues: for \(P=2I_n\) the new Gram matrix is \(4A\).
See [@Art11].