2

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 $\begin{bmatrix}A\end{bmatrix} \begin{bmatrix} v_1 & v_2 \end{bmatrix} = \begin{bmatrix} Av_1 & Av_2 \end{bmatrix}$?

user35603
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committedandroider
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1 Answers1

2

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 $AB = A[\mathbf{v}\; \mathbf{w}] = [A\mathbf{v}\; A\mathbf{w}]$.

Fargle
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