I know it's probably a silly question, but I'm trying to figure out why was matrix multiplication (the standard one) defined the way it was defined.
I know that it was defined like that so we would gain invariance under change of basis: $PAP^{-1}+PBP^{-1}=P(A+B)P^{-1}$ and $(PAP^{-1})(PBP^{-1})=PABP^{-1}$ and that ofcourse the case.
But another explanation that was suggested is: "We defined matrix multiplication this way so that if $A$ is the matrix of a linear transformation $T_1$ with respect to some basis $s$ and $B$ is the matrix of a linear transformation $T_2$ with respect to the same basis $s$ then $AB$ is the matrix of $T_1$ composition with $T_2$ (I don't know the command for composition operator) with respect to basis $s$.
Again, this is a completely legitimate aspiration, but I fail to see why it follows. Why is linear transformation composition equivalent to matrix multiplication?
Best Answer
Hint:
test that: $$ A(B\vec x)=(AB) \vec x $$ Using for $\vec x$ the canonical basis. It is easy for transformations in a $2$ dimensional space and requaire a bit more work for an $n-$ dimensionale space.
Given $$ A=\begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix} \qquad B=\begin{bmatrix} b_{11}&b_{12}\\ b_{21}&b_{22} \end{bmatrix} $$ By row-column multiplication we have: $$ A(B\vec x)= \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix} \left( \begin{bmatrix} b_{11}&b_{12}\\ b_{21}&b_{22} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \right)= \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix} \begin{bmatrix} b_{11}x+b_{12}y\\ b_{21}x+b_{22}y \end{bmatrix}= $$ $$ = \begin{bmatrix} a_{11}(b_{11}x+b_{12}y)+a_{12}(b_{21}x+b_{22}y)\\ a_{21}(b_{11}x+b_{12}y)+a_{22}(b_{21}x+b_{22}y) \end{bmatrix} $$ That, reordering becomes: $$ \begin{bmatrix} (a_{11}b_{11}+a_{12}b_{21})x+(a_{11}b_{12}+a_{12}b_{22})y\\ (a_{21}b_{11}+a_{22}b_{21})x+(a_{21}b_{12}+a_{22}b_{22})y \end{bmatrix}=(AB)\vec x $$