# 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.

• Can you give an instructive example? Sep 4, 2015 at 15:19
• The operation looks very similar to the tensor product, but I don't fully understand the last two sentences.. Sep 4, 2015 at 19:14

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.