This is a question that has been bugging me for some time. I know that it's true, but up till now I'm still not sure how to prove it.
Let A be an m x n matrix, and let X be an n x p matrix. Let ($x_1 x_2 … x_p$) be the columns of X. Prove that AX = ($Ax_1 Ax_2 … Ax_p$).
I understand that AX = A($x_1 x_2 … x_p$). Why is it, then, that we can casually bring A into the parenthesis such that A($x_1 x_2 … x_p$) = ($Ax_1 Ax_2 … Ax_p$)?
EDIT: After much thought, I have devised a proof.
(i,j)-entry of AX = (i,j)-entry of A($x_1 x_2 … x_p$) = $\sum_{k=1}^n a_i$$_k.x_k$$_j$
$j^t$$^h$ column of ($Ax_1 Ax_2 … Ax_p$) = $Ax_j$ and within $Ax_j$ lies the (i,j)-entry of ($Ax_1 Ax_2 … Ax_p$).
Hence (i,j)-entry of ($Ax_1 Ax_2 … Ax_p$) = $R_ix_j$ where $R_i$ is row i of A = $\sum_{k=1}^n a_i$$_k.x_k$$_j$
Since the size of AX and ($Ax_1 Ax_2 … Ax_p$) are both m x p and the (i,j)-entry of AX = (i,j) entry of ($Ax_1 Ax_2 … Ax_p$), I conclude that AX = ($Ax_1 Ax_2 … Ax_p$).
Is this proof valid?
Best Answer
This is just a (well known) formal property of matrix multiplication. From the definition, the $(i,j)$ entry of the matrix product $AB$ is the same as the unique entry of the product $R_iC_j$ where $R_i$ is row $i$ of $A$ (a $1\times n$ matrix) and $C_j$ is column $j$ of $B$ (a $n\times 1$ matrix). By varying $i$ it follows from this that column $j$ of $AB$ equals $AC_j$, in other words each column of $AB$ can be computed using the corresponding column of $B$ only. This answers you question. Alternatively one can vary $j$ to find that row $i$ of $AB$ equals $R_iB$, so each row of $AB$ can be computed using the corresponding row of $A$ only. Both points of view are occasionally useful.