# Matrix Multiplication part 2

This is part two of the “boring stuff.” I put that in quotes to evoke a sense of perspective. How boring can it be? We’re still doing math.

Suppose I have linear transformations $T:U\to V$, and $S:V\to W$. Let $\mathcal A$, $\mathcal B$, and $\mathcal C$ be bases for $U$, $V$, and $W$ respectively. What happens when I compose $T$ and $S$? I sure hope that it turns out to me multiplying the matrices $(T)_{\mathcal A\to \mathcal B}$ and $(S)_{\mathcal B\to \mathcal C}$.

Let $\mathcal A=\{a_1,\dots,a_\ell\}$$\mathcal B=\{b_1,\dots,b_m\}$$\mathcal C=\{c_1,\dots,c_n\}$.

Let $M=(S\circ T)_{\mathcal A\to \mathcal C}$. I want to know what the entry $M_{i,j}$ in the matrix is. So I want to find out what $S\circ T$ does to a the basis vector $a_i$ and then look at the component of $c_j$. That will give me the appropriate coefficient.

$(S\circ T)a_i=S(Ta_i)=S\left(\displaystyle\sum_{k=1}^mt_{k,i}b_i\right)=\displaystyle\sum_{k=1}^mt_{i,k}S(b_k)=\sum_{k=1}^mt_{i,k}\left(\sum_{p=1}^ns_{k,p}c_p\right)$

Of course, we only care about the coefficient of $c_j$, so we can disregard any time $p\neq j$. This gives us:

$M_{i,j}=\displaystyle\sum_{k=1}^mt_{i,k}s_{k,j}$

And of course, this is exactly what you would get if you multiplied matrices $(S_{\mathcal B\to\mathcal C})\cdot( T_{\mathcal A\to\mathcal B})$ the way you were taught.

Okay. I’m pretty sure we’re done with annoying computations for a while.