8
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Thanks a ton, that makes sense. $\endgroup$
    – moose
    Sep 15 '11 at 1:35
  • $\begingroup$ Mr. Strang is the way to go. $\endgroup$
    – marnix
    Nov 6 '21 at 13:17
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.