Let \(V\) be a finite-dimensional inner product space, and let \(\mathfrak{B} = \{\alpha_1, \ldots, \alpha_n\}\) be an orthonormal basis for \(V\). Let \(T\) be a linear operator on \(V\) and \(A\) the matrix of \(T\) in the ordered basis \(\mathfrak{B}\). Prove that \[ A_{ij} = (T\alpha_j \mid \alpha_i). \]
Solution
By definition of the matrix \(A=[T]_{\mathfrak B}\), its \(j\)th column is the coordinate vector of \(T\alpha_j\). Thus \[ T\alpha_j=\sum_{k=1}^n A_{kj}\alpha_k. \] Take the inner product with \(\alpha_i\). Using orthonormality, \[ \begin{aligned} (T\alpha_j\mid\alpha_i) &=\sum_{k=1}^n A_{kj}(\alpha_k\mid\alpha_i)\\ &=\sum_{k=1}^n A_{kj}\delta_{ki}\\ &=A_{ij}. \end{aligned} \] Therefore \[ \boxed{A_{ij}=(T\alpha_j\mid\alpha_i)} \] for all \(i,j\) [@HK71].