Suppose $C = AB$. Show $\hat{c}_j = \sum_{k} b_{kj}\hat{a}_{k}$.

Question

Suppose $C = AB$. Show $\hat{c}_j = \sum_{k} b_{kj}\hat{a}_{k}$.

$A$, $B$, and $C$ are square matrices of the same size.

$\hat{c}_j$ is the $j$th column of $C$, $\hat{a}_k$ are the columns of $A$, $b_{kj}$ are the entries of the $j$th column of $B$.

I was given this hint:

What is the $i$th column of each side?

I guess I'm just having a hard time understanding the meaning of all of this. I realize that I can multiply $A$ and $B$ and obtain different columns of $C$, which I tested with 2x2 matrices, but nothing stood out to me. I'm not sure why finding another column of both sides is going to help me with the $j$th column.

Answer 1

We know that the $j-th$ column of $AB$ is given by: $$\hat c_{j} \begin{array}[t]{l}= A\cdot \hat b_j =\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \cdot \begin{bmatrix} b_{1j} \\ b_{2j}\\\vdots \\ b_{nj} \end{bmatrix} = \begin{bmatrix} a_{11} b_{1j} + a_{12}b_{2j} + \cdots + a_{1n}b_{nj}\\ a_{21} b_{1j} + a_{22}b_{2j} + \cdots + a_{2n}b_{nj}\\ \vdots\\ a_{n1}b_{1j} + a_{n2} b_{2j} + \cdots + a_{nn}b_{nj} \end{bmatrix}\\\\ = b_{1j}\begin{bmatrix} a_{11}\\a_{21}\\\vdots\\ a_{n1}\end{bmatrix} + b_{2j} \cdot \begin{bmatrix} a_{12}\\a_{22}\\\vdots\\ a_{n2}\end{bmatrix}+\cdots+ b_{nj}\cdot \begin{bmatrix} a_{1n}\\a_{2n}\\\vdots\\ a_{nn}\end{bmatrix}. \end{array}$$ The result follows.