Matrix multiplication of columns times rows instead of rows times columns In ordinary matrix multiplication $AB$ where we multiply each column $b_{i}$ by $A$, each resulting column of $AB$ can be viewed as a linear combination of $A$. 
If however if we decided to multiply each column of $A$ by each row of $B$, we  get an entire matrix for each column-row multiply. My question is: Does each matrix resulting from an outer product have any known meaning aside from being a part of the sum(a summand?) of the final $AB$?
Edit: Say we have $AB$
$$ \left( \begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc}
10 & 11 & 12 \\
13 & 14 & 15 \\
15 & 16 & 17 \end{array} \right) $$
Normally we would multiply each column of $B$ by A and get a linear combination of A, e.g.
$$10\left( \begin{array}{c}
1 \\ 4\\ 7 \end{array} \right)+ 13\left( \begin{array}{c}
2 \\ 5\\ 8 \end{array} \right)+ 15\left( \begin{array}{c}
3 \\ 6\\ 9 \end{array} \right)$$ which is one column of $AB$.
If however we multiply each column of $A$ by each row of $B$, e.g.
$$\left( \begin{array}{c}
1 \\ 4\\ 7 \end{array} \right)\left( \begin{array}{ccc}
10 & 11 & 12 \end{array} \right)$$ we get a matrix. Each of the 3 matrices $a_{i}b_{i}^{T}$ summed together gives us $AB$. I was wondering if each individual matrix that sums to $AB$ has any sort of special meaning. This second way of performing multiplication also seems to be called column-row expansion. (http://www.math.nyu.edu/~neylon/linalgfall04/project1/dj/crexpansion.htm). I actually read about it in I believe section 2.4 of Strang's Introduction to Linear Algebra book. He mentions that not everybody is aware that matrix multiplication can be performed in this way.
 A: 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. 
