Let \(V\) and \(W\) be finite-dimensional vector spaces over the field \(F\) and let \(T\) be a linear transformation from \(V\) into \(W\). If \[ \mathcal{B} = \{\alpha_1, \ldots, \alpha_n\} \quad \text{and} \quad \mathcal{B}' = \{\beta_1, \ldots, \beta_m\} \] are ordered bases for \(V\) and \(W\), respectively, define the linear transformations \(E^{p,q}\) as in the proof of Theorem 5: \(E^{p,q}(\alpha_i) = \delta_{iq}\beta_p\). Then the \(E^{p,q}\), \(1 \leq p \leq m\), \(1 \leq q \leq n\), form a basis for \(L(V, W)\), and so \[ T = \sum_{p=1}^{m} \sum_{q=1}^{n} A_{pq} E^{p,q} \] for certain scalars \(A_{pq}\) (the coordinates of \(T\) in this basis for \(L(V, W)\)). Show that the matrix \(A\) with entries \(A(p, q) = A_{pq}\) is precisely the matrix of \(T\) relative to the pair \(\mathfrak{B}, \mathfrak{B}'\).
Solution
For each basis vector \(\alpha_q\), \[ T\alpha_q =\sum_{p=1}^m A_{pq}E^{p,q}(\alpha_q) +\sum_{p=1}^m\sum_{j\ne q}A_{pj}E^{p,j}(\alpha_q). \] By definition, \[ E^{p,q}(\alpha_q)=\beta_p, \qquad E^{p,j}(\alpha_q)=0\quad(j\ne q). \] Therefore \[ T\alpha_q=\sum_{p=1}^m A_{pq}\beta_p. \] Thus the coordinate column of \(T\alpha_q\) in the ordered basis \(\mathfrak B'\) is exactly \[ (A_{1q},\ldots,A_{mq})^t. \] These are precisely the columns of the matrix of \(T\) relative to the pair \(\mathfrak B,\mathfrak B'\). Hence that matrix is \(A=(A_{pq})\).