Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there some standard operation to "rotate rings on matrices"? Look at the image below:

The numbers around the four empty squares are what I'm calling ring, In the second matrix, this ring has been rotated counterclockwise. I'm aware that ring may not be the right name.

share|improve this question
    
What should I review? –  Vÿska Jul 5 '12 at 7:38
    
Ring is perhaps a poor choice of words, since it's very common to talk about "matrices defined over a ring", which has a very different meaning. –  Alex Becker Jul 5 '12 at 7:39
    
Oddly enough, if you do the same thing to $2\times 2$ matrices and restrict you attention to those of the form $$\begin{pmatrix}a & b \\ c & a\end{pmatrix}$$ then rotation counterclockwise corresponds to multiplication on the right by $$\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$$ while rotation clockwise corresponds to multiplication on the left by the same matrix. –  Alex Becker Jul 5 '12 at 8:07
    
@AlexBecker Yep, But I've explained it. It's also because my native languange is portuguese, I just didn't know which word to use. Suggestions? –  Vÿska Jul 5 '12 at 9:23

3 Answers 3

up vote 3 down vote accepted

Use the vector representation $\mathrm{vec}$ of your matrix $X$, by stacking all columns on top of each other: $$ \mathrm{vec}\: \pmatrix{ 1 &4&2&6\\ 7&\square&\square&2 \\ 7&\square&\square&4 \\ 8&2&5&6\\ } =\pmatrix{ 1 &4&2&6& 7&\square&\square&2 & 7&\square&\square&4 & 8&2&5&6\\ }^T. $$ Now apply $\pi_{\text{rot.ring}}$, a permutation (matrix $M$, with dimension $4^2$) on $\mathrm{vec}\;X$, such that $\left(\mathrm{vec}\;X\right)_1$ is sent to $\left(\mathrm{vec}\;X\right)_5$, $\left(\mathrm{vec}\;X\right)_2$ to $\left(\mathrm{vec}\;X\right)_1$,... So $\pi_{\text{rot.ring}}=\left(5,1,2,3,4,8,12,16,15,14,13,9\right)$ and $M_{j,j+1}=\left(\pi_{\text{rot.ring}}\right)_{j,j+1}$ and $M_{jj}=1$ if $j\in\{6,7,10,11\}$.

Undo the $\mathrm{vec}$ operation and you'll get a matrix with your ring rotated.

share|improve this answer
    
Will this procedure work on matrices bigger than 4x4? And will the same procedure work for rotating the inner "rings"? –  Vÿska Jul 5 '12 at 22:42
    
Sure, this will work for larger matrices and other rings (for the inner ring cycle the elements $\{6,7,10,11\}$). –  draks ... Jul 6 '12 at 6:32
    
@GustavoBandeira do you agree? –  draks ... Jul 11 '12 at 11:05
    
Actually I'll need to translate what you said - and also what others said - so I can apply it. I was just ashamed of asking for help but here it goes: What topics do I need to learn? I supose matrices are the topic, but could it be more specifical? –  Vÿska Jul 11 '12 at 13:27
    
Are you asking for a good book on matrices? Here you'll find some examples... –  draks ... Jul 11 '12 at 13:46

I'm pretty sure this operation has absolutely no interesting interpretation when the matrix is viewed as representing a linear transformation, as matrices usually are. It is therefore a fairly certain bet that this operation has no name.

share|improve this answer
    
I believed it has no name too, but I'm naive then I decided to ask. –  Vÿska Jul 5 '12 at 22:39

If you identify the space $M_{n\times n}$ with $\mathbb R^{n^2}$, and let $S_{n^2}$ act on $\mathbb R^{n^2}$ by permuting the axes, then your operation corresponds to the action of a $4n-4$ cycle on $M_{n\times n}$. This interpretation is purely geometrical though, and does not give any understanding of how the operation affects the linear transformation corresponding to a matrix.

share|improve this answer
    
Will this procedure work on matrices bigger than 4x4? And will the same procedure work for rotating the inner "rings"? –  Vÿska Jul 5 '12 at 22:41
    
@GustavoBandeira Yes, it works for any $n$. And any "ring" as well,. However, you must choose the correct $4n-4$ cycle; not just any cycle will do. –  Alex Becker Jul 6 '12 at 1:03
    
What you mean with "cycle", is this a name for my "ring"? –  Vÿska Jul 6 '12 at 2:39
    
@GustavoBandeira No, "cycle" is a technical term. It describes certain elements of the permutation group $S_{n^2}$. You can read about cycles on Wikipedia. –  Alex Becker Jul 6 '12 at 2:41
    
Oh, ok. I was confused because you suggested a better term instead of "ring". –  Vÿska Jul 6 '12 at 2:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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