# Matrix Multiplication

Warning: Today and tomorrow are going to be painful. We’re checking that multiplication of matrices and vectors the way we learned works the way we it should. It’s boring, but necessary.

If I have a linear transformation $T$ from $V$ to $W$, and $\mathcal B=\{v_1,\dots,v_n\}$ is a basis for $V$ and $\mathcal C=\{w_1,\dots,w_m\}$ is a basis for $W$, I can write the matrix $T_{\mathcal B\to\mathcal C}$. If I have a vector $v\in V$, I can write that in its “column notation” in the basis $\mathcal B$. Let’s say

$v=\lambda_1v_1+\cdots+\lambda_nv_n=\left(\begin{array}{c}\lambda_1\\ \vdots \\ \lambda_n\end{array}\right)_\mathcal B$,

and say that

$Tv_i=a_{1,i}w_1+\cdots+a_{m,i}w_m=\left(\begin{array}{c}a_{1,i}\\ \vdots \\ a_{m,i}\end{array}\right)_\mathcal C$.

Then we can use the linearity of $T$ to determine $Tv$ in terms of the $Tv_i$.

$Tv = T\left(\displaystyle\sum_{i=1}^n\lambda_i v_i\right)=\displaystyle\sum_{i=1}^nT(\lambda_i v_i)=\displaystyle\sum_{i=1}^n\lambda_i Tv_i$

but then we know each $Tv_i$ in terms of the $w_j$, so we can figure out the entire sum in terms of the $w_j$.

$Tv=\displaystyle\sum_{i=1}^n\lambda_i\left(\displaystyle\sum_{j=1}^ma_{j,i}w_j\right)=\sum_{j=1}^m(a_{1,1}\lambda_1+\cdots+a_{1,n}\lambda_n)w_j$

If we write this all in the “column notation,” it would say:

$\left(\begin{array}{ccc}a_{1,1} & \cdots & a_{1,n}\\ \vdots & \ddots & \vdots\\ a_{m,1} & \cdots & a_{m,n}\end{array}\right)_{\mathcal B\to\mathcal C}\left(\begin{array}{c}\lambda_1\\ \vdots\\ \lambda_n\end{array}\right)_{\mathcal B}=\left(\begin{array}{c}a_{1,1}\lambda_1+\cdots+a_{1,n}\lambda_n\\ \vdots\\ a_{m,1}\lambda_1+\cdots+a_{m,n}\lambda_n\end{array}\right)_{\mathcal C}$

You may notice that this is exactly the way you probably learned to multiply a vector by a matrix. What does this mean? It means back in the day, no one told you why you did it that way, but now you know why its correct.