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Usual matrix multiplication is done from left to right and top to bottom. Does there exist an application or a theory that does matrix multiplication from right to left and top to bottom?

EXAMPLE:

$\begin{bmatrix}0 \ 1 \\ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \\ 1 \ 0 \end{bmatrix} « \begin{bmatrix} 0 \ 1 \\ 1 \ 0 \end{bmatrix}$

Where I have used the symbol « to denote that the multiplication must be done strictly from RIGHT to LEFT and TOP to BOTTOM.

To get the $a_{1,2} $ entry of the product we would do $ 1 \cdot 1 + 0 \cdot 0 = 1 $

In words,

(first row of matrix on the right of «)×(second column of matrix on the left of «) = (the entry in the first row , second column of the product)

The other entries of the product are computed in the same way. Has this been explored by anyone? Is there any published work?

Thank you for your consideration in this matter.

neofoxmulder
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  • Could you write down the exact definition of $«(A,B)$ please? – user127.0.0.1 Jan 16 '14 at 04:52
  • $\begin{bmatrix}fa + ec \ fb + ed \ ha + gc \ hb + gd \end{bmatrix} = \begin{bmatrix}a \ b \ c \ d \end{bmatrix} « \begin{bmatrix}e \ f \ g \ h \end{bmatrix}$

    Should i have edited my original post or add it as a comment? Note i can't preview MathJax for comments. Thanx.

     – neofoxmulder Jan 16 '14 at 05:02

2 Answers2

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It doesn't seem like there's anything here that can't be done with every-day matrix multiplication. Seems like we can calculate $A«B$ as follows:

  1. flip $A$, call the flipped version $A_F$. Same for $B_F$
  2. Calculate $B_FA_F$
  3. Take the product, flip it back

As it ends up, "flipping" a matrix from right to left is itself a matrix operation. Namely, let $K_n$ be the $n\times n$ matrix given by

$$ K_n = \pmatrix{ 0&\cdots&0&1\\ \vdots&&1&0\\ 0&& &\vdots\\ 1&0&\cdots &0 } $$

Now, suppose $A$ is $k \times m$ and $B$ is $n \times k$. Then we can calculate $$ A«B= (B K_k)(AK_m)K_m = BK_kA $$ I have not heard of any application for this particular set of operations.

This analysis also reveals two different ways of finding $A«B$:

  • flip $B$ right to left to get $B_F$, and calculate $A«B = B_F A$
  • flip $A$ top to bottom to get $A_F$, and calculate $A«B = B A_F$
Ben Grossmann
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  • Hey! That's very helpful , Thanx! I don't have enough rep to vote you up yet. I'm trying to find easy 'square roots' of matrices and discover when and how many such 'square roots' exist. My interest in this matter began when considering

    $ A^2 = \begin{bmatrix} 0 \ 1 \ 1 \ 0 \end{bmatrix} $

     – neofoxmulder Jan 16 '14 at 06:07
  • Do you mean answers to $A\cdot A = \pmatrix{0&1\1&0}$ under typical matrix multiplication, or do you mean $A«A = \pmatrix{0&1\1&0}$? – Ben Grossmann Jan 16 '14 at 13:51
  • Also, if you do mean the latter, are you aware of the results for the existence of square roots under the usual matrix multiplication? That can get a bit tricky. And are you including matrices with complex-valued entries? – Ben Grossmann Jan 16 '14 at 14:05
  • AFIK , if we restrict to real numbers and usual matrix multiplication then $ A \cdot A = \begin{bmatrix} 0 \ 1 \ 1 \ 0 \end{bmatrix} $ does not have a solution. Under the « operator it does , $ A = \begin{bmatrix} 0 \ 1 \ 1 \ 0 \end{bmatrix} $ It is it's own square root , sort of like $\sqrt{1} = 1 $ or $ \sqrt{0} = 0 $ I don't know if there are other square roots as well.

    If we allow complex entries then usual matrix multiplication $ A \cdot A = \begin{bmatrix} 0 \ 1 \ 1 \ 0 \end{bmatrix} $ has 4 distinct rather 'nasty' square roots , AFAIK.

    Thanx again for your replies in this matter.

     – neofoxmulder Jan 16 '14 at 18:53
  • The square roots of $\pmatrix{0&1\1&0}$ under $«$ are just going to be the square roots of $\pmatrix{1 & 0\0&1}$ (the identity matrix), but reversed. In this case, you also get $4$ square roots, which are $$A = \pmatrix{\pm 1 & 0\ 0 & \pm 1}$$ Reverse that to get the square roots of $A$ under $«$. – Ben Grossmann Jan 16 '14 at 23:43
  • Thanx for the valuable feedback. Please note that under normal multiplication $ A = \begin{bmatrix} 0 \ 1 \ 1 \ 0 \end{bmatrix} $ is also a square root of $\begin{bmatrix} 1 \ 0 \ 0 \ 1 \end{bmatrix} $. Under typical matrix multiplication , matrices of the form $ \begin{bmatrix} k \ 0 \ 0 \ k \end{bmatrix} $ have an infinite number of square roots $ A = \begin{bmatrix}\sqrt{k} \ \ \ \ \ \ c \ \ \ 0 \ - \sqrt{k} \end{bmatrix} $. There are other solutions as well. If you have the time , look at this thread,http://www.mymathforum.com/viewtopic.php?f=18&t=45437 where this adventure began. – neofoxmulder Jan 17 '14 at 07:30
  • @neofoxmulder right, my mistake. In face, any matrix of the form $S\pmatrix{1 & 0\0 & -1}S^{-1}$, where $S$ is an invertible matrix, will be a square root of the identity matrix. The key to this kind of analysis comes down to eigenvalues, an perhaps Jordan canonical form. I think that's true for square roots in general – Ben Grossmann Jan 17 '14 at 13:20
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Don't know of any applications or previous work, but your operation can be expressed in terms of ordinary matrix multiplication, at least in the $2\times2$ case: $$A\,«\,B=BJA$$ where $$J=\begin{bmatrix} 0 \ 1 \\ 1 \ 0 \end{bmatrix}\ .$$ It is then easy to show that your operation is associative, distributive over addition, etc. Moreover, because $J$ is invertible, any $2\times2$ matrix can be written in the form $XJ$, and your definition can be written $$(YJ)\,«\,(XJ)=XYJ\ .$$ What this comes down to is that the equation $$\phi(X)=XJ$$ defines a bijection of $M_{2,2}$ onto itself (in informal language, it's just a relabelling of the elements, with no essential change in properties) which preserves addition, $$\phi(X+Y)=\phi(X)+\phi(Y)\ ,$$ and converts multiplication to your operation, with change of order, $$\phi(XY)=\phi(Y)\,«\,\phi(X)\ .$$ You can then find an identity for your operation, and an inverse (when it exists).

Regardless of any applications, I think that this would be a very good project for students beginning to work in abstract algebra. Thanks for bringing it to my attention!

David
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  • Also very helpful, Thanx!

    As an example ,

    $ \begin{bmatrix}5 \ 8 \ 13 \ 20 \end{bmatrix} = \begin{bmatrix}1 \ 2 \ 3 \ 4 \end{bmatrix} « \begin{bmatrix}1 \ 2 \ 3 \ 4 \end{bmatrix}$

    So in a 'some sense' kind of way , $\begin{bmatrix}5 \ 8 \ 13 \ 20 \end{bmatrix}$ has an easy 'square root' under the « operation.

     – neofoxmulder Jan 16 '14 at 06:16