Matrix Multiplication

Let \(A\) and \(B\) be \(m \times n\) matrices. If \(A \boldsymbol{x} = B \boldsymbol{x}\) for all \(\boldsymbol{x} \in \mathbb{R}^n\), then \(A = B\).

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Proof. The standard basis of \(\mathbb{R}^n\) is

\[\begin{split} \boldsymbol{e}_1 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \ \ , \ \ \boldsymbol{e}_2 = \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix} \ \ , \ \ \dots \ \ , \ \ \boldsymbol{e}_n = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix} \end{split}\]

In other words, \(\boldsymbol{e}_k\) is the vector with 1 at index \(k\) and 0 everywhere else. Then \(A \boldsymbol{e}_k\) is equal to the \(k\)th column of \(A\). Since \(A \boldsymbol{e}_k = B \boldsymbol{e}_k\) for each \(k=1,\dots,n\) we see that the columns of \(A\) and \(B\) are equal therefore \(A = B\).

Let \(A\) be a \(p \times m\) matrix, let \(B\) be a \(m \times n\) matrices and let \(\boldsymbol{b}_1,\dots,\boldsymbol{b}_n\) be the columns of \(B\)

\[\begin{split} B = \begin{bmatrix} & & \\ \boldsymbol{b}_1 & \cdots & \boldsymbol{b}_n \\ & & \end{bmatrix} \end{split}\]

Then the \(k\)th column of \(AB\) is \(A \boldsymbol{b}_k\). In other words, matrix multiplication \(AB\) can be written as

\[\begin{split} AB = \begin{bmatrix} & & \\ A \boldsymbol{b}_1 & \cdots & A \boldsymbol{b}_n \\ & & \end{bmatrix} \end{split}\]