Question

Show that, for each k = 1,...,K, Col_k(C), the kth column of the matrix C = AB, is

Col_k(C) = ACol_k(B).

Answer 1

Question 1.

Show that for each $k=1,\ldots,K$, $Col_{k}(C)$, the $k$-th column of the matrix $C=AB$, is $Col_{k}(C)=ACol_{k}(B)$.

Solution. Recall that $Col_{k}(C)$ consists of the elements $C(i,k)$ of the matrix $C$, $i=1,\ldots,K$. By definition of the product of matrices:

$C(i,k)=(AB)(i,k)=\sum_{j=1}^{K}A(i,j)B(j,k).$

Thus, $C(i,k)$ is obtained by “multiplication” of the $i$-th row of $A$ by the $k$-th column $Col_{k}(B)$ of $B$. Therefore, $Col_{k}(C)$ is the result of applying $A$ to the vector $Col_{k}(B)$. $\Box$