Let \(A\) be an \(m \times n\) matrix and \(B\) an \(n \times k\) matrix. Show that the columns of \(C = AB\) are linear combinations of the columns of \(A\). If \(\alpha_1, \ldots, \alpha_n\) are the columns of \(A\) and \(\gamma_1, \ldots, \gamma_k\) are the columns of \(C\), then \[ \gamma_j = \sum_{r=1}^{n} B_{rj}\alpha_r. \]
Solution
Fix \(1\le j\le k\). The \(i\)th entry of the \(j\)th column \(\gamma_j\) of \(C=AB\) is \[ C_{ij}=\sum_{r=1}^n A_{ir}B_{rj}. \] On the other hand, the \(i\)th entry of \[ \sum_{r=1}^n B_{rj}\alpha_r \] is the same sum, since the \(i\)th entry of \(\alpha_r\) is \(A_{ir}\). Hence \[ \gamma_j=\sum_{r=1}^n B_{rj}\alpha_r. \] Thus every column of \(AB\) is a linear combination of the columns of \(A\).