2
$\begingroup$

Basically

My professor wrote down simplified this expression

$Av_1 =$

$Av_2 =$

to

$\begin{bmatrix}A\end{bmatrix} \begin{bmatrix} v_1 & v_2 \end{bmatrix}$

Where $A$ is a matrix, $v_1$ and $v_2$ are vectors. Why can this be simplified down this way. The way I do matrix multiplication is to multiply the row by column. If A is a two by two matrix, would you do the proof by pretending $A$ is a constant and saying that $\begin{bmatrix}A\end{bmatrix} \begin{bmatrix} v_1 & v_2 \end{bmatrix} = \begin{bmatrix} Av_1 & Av_2 \end{bmatrix}$?

$\endgroup$
2
$\begingroup$

Pretending $A$ is a constant is not how the proof is done, but it is one way to think about it.

The proof is achieved by using the definition of matrix multiplication:

Let $a_{ij}$ be the element in the $i$th row and $j$th column of $m \times n$ matrix $A$, and let $\mathbf{v}$ and $\mathbf{w}$ be two column $n$-vectors with coordinates $v_j$ and $w_j$. Then the $i$th element of $A\mathbf{v}$ is $\sum_{k} a_{ik}v_k$; similarly for $A\mathbf{w}$.

Now consider the matrix $B = [\mathbf{v}\; \mathbf{w}]$, and let $b_{ij}$ denote the element at the $i$th row and $j$th column. The $i$th element of the first column of the product $AB$ will be $\sum_{k}a_{ik}b_{k1}$, and that of the second column will be $\sum_{k}a_{ik}b_{k2}$. (Note that there can be no more columns, as an $m \times n$ matrix times an $n \times 2$ matrix will be an $m \times 2$ matrix.)

But $b_{k1}$ is exactly $v_k$, and $b_{k2}$ is exactly $w_k$. Therefore, we have that the first column of $AB$'s elements are $\sum_{k} a_{ik}v_k = (A\mathbf{v})_i$, and the second column's elements are $(\sum_{k} a_{ik}w_k = (A\mathbf{w})_i$, so that $AB = A[\mathbf{v}\; \mathbf{w}] = [A\mathbf{v}\; A\mathbf{w}]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.