Matrix multiplication of columns times rows instead of rows times columns

Before talking about multiplication of two matrices, let's see another way to interpret matrix $A$. Say we have a matrix $A$ as below, $$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} $$ we can easily find that column $\begin{bmatrix} 3 \\ 2 \\ 3 \\\end{bmatrix}$ is linear combination of first two columns. $$ 1\begin{bmatrix} 1 \\ 1 \\ 1\\\end{bmatrix} + 1\begin{bmatrix} 2 \\ 1 \\ 2\\\end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 3 \\\end{bmatrix} $$ And you can say $\begin{bmatrix} 1 \\ 1 \\ 1 \\\end{bmatrix}$ and $\begin{bmatrix} 2 \\ 1 \\ 2 \\\end{bmatrix}$ are two basis for column space of $A$.

Forgive the reason why you want to decompose matrix $A$ at first place like this, $$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \\ \end{bmatrix} + \begin{bmatrix} 0 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & 2 & 2 \\ \end{bmatrix} $$ but you can, and in the end, it looks reasonable.

If you view this equation column wise, each $column_j$ of $A$ is the sum of corresponding $column_j$ of each matrix in RHS.

What's special about each matrix of RHS is that each of them is a rank 1 matrix whose column space is the line each base of column space of $A$ lies on. e,g. $ \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \\ \end{bmatrix} $ spans only $\begin{bmatrix} 1 \\ 1 \\ 1 \\\end{bmatrix}$. And people say rank 1 matrices are the building blocks of any matrices.

If now you revisit the concept of viewing $A$ column by column, this decomposition actually emphasizes the concept of linear combination of base vectors.

If these make sense, you could extend the RHS further, $$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \\\end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\\end{bmatrix} + \begin{bmatrix} 2 \\ 1 \\ 2 \\\end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\\end{bmatrix} $$ Each term in RHS says take this base, and make it "look like" a rank 3 matrix.

And we can massage it a little bit, namely put RHS into matrix form, you get $$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & 1 \\ 1 & 2 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{bmatrix} $$

Now you can forget matrix $A$, and imagine what you have are just two matrices on RHS. When you read this text backward(I mean logically), I hope matrix multiplication in this fashion makes sense to you now. Or if you prefer, you can start with two matrices in the question.