matrix multiplication by columns

Question

Yo,

I need some help with understanding matrix multiplication by columns. Consider the two matrices:

$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \\ 7 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} 3 & 2 & 1 \\ 4 & 5 & 6 \\ 9 & 8 & 7 \end{array} \right) $

So I'm familiar with the standard algorithm where element $AB_{ij}$ is found by multiplying the $i^{th}$ row of A with the $j^{th}$ column of B.

Apparently there is another way to multiply matrices where you work with whole columns of A to get the product AB. Does anyone know how to do that? If so, could you please provide a general algorithm? I've never heard of it and I can't find it anywhere.

score 11 · Accepted Answer · answered Sep 15 '11 at 01:03

11

You can find it in the starting lectures of Gilbert Strang. Anyway, here is just a teaser for you $$ \left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \\ 7 & 8 & 9 \end{array} \right)\begin{pmatrix}x\\y\\z \end{pmatrix} = \begin{pmatrix} 1\\6\\7 \end{pmatrix}x+ \begin{pmatrix} 2\\5\\8 \end{pmatrix}y+ \begin{pmatrix} 3\\4\\9 \end{pmatrix}z $$ How 'bout that?

answered Sep 15 '11 at 01:03

Thanks a ton, that makes sense. – moose Sep 15 '11 at 01:35
Mr. Strang is the way to go. – marnix Nov 06 '21 at 13:17

score 0 · Answer 2 · answered Nov 06 '21 at 12:44

Not an answer, but related: you can do the same thing, but with rows of a matrix:

$$\begin{pmatrix} x & y & z \end{pmatrix} \cdot \left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \\ 7 & 8 & 9 \end{array} \right) = x\cdot \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} + y\cdot \begin{pmatrix} 6 & 5 & 4 \end{pmatrix} + z\cdot \begin{pmatrix} 7 & 8 & 9 \end{pmatrix}.$$

This is not a surprise, as you can just transpose the product and use the column method to compute the product. After that, you transpose the result again.

Still think it is worth a share.

matrix multiplication by columns

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