# Matrix doodles

Discussion in 'Physics & Math' started by arfa brane, Sep 18, 2016.

1. ### arfa branecall me arfValued Senior Member

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Suppose you'd like to map the rotation group of a square to a square matrix.

First of all a 2 x 2 matrix should suffice to describe the four vertices, i.e. something like:

$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$
or if you want to represent a 4-cycle: (clockwise labeling):

$\begin{pmatrix} a & b \\ d & c \end{pmatrix}$
The permutation matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, acting on the left of any 2 x 2 matrix will swap the rows. Acting on the right it will swap the columns.

$\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ d & c \end{pmatrix} = \begin{pmatrix} d & c \\ a & b \end{pmatrix}$

$\begin{pmatrix} a & b \\ d & c \end{pmatrix} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}b & a \\ c & d \end{pmatrix}$
So with the matrix transpose, a rotation or cyclic permutation of the elements of a 2 x 2 matrix can be defined:

$\begin{pmatrix} a & b \\ d & c \end{pmatrix}^{\intercal}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}d & a \\ c & b \end{pmatrix} = \biggl {(} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ d & c \end{pmatrix} \biggr {)}^{\intercal}$
But you can also map the four vertices {a,b,c,d} to a column vector:

$\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}$​

Then rotations are represented by 4 x 4 matrices, a subset of the 24 possible permutations of four symmetric objects. Can the above representation of rotating a square be extended to the whole of $S_4$, which is the symmetry group of a solid cube ? A cubic matrix will be needed, and a way to define rotations of the entire matrix (or equivalently, rotations of an abstract system of coordinates, given by a set of indices).

$\begin{pmatrix} e & f \\ h & g \end{pmatrix}$
With connections between the rows/columns of this matrix to the first one. Index-wise, this is just a pair of 2 x 2 matrices connected, or 'sliced' as $M_{1ij} + M_{2ij}$. Obviously the indices will allow three distinct slices of the 2 x 2 x 2 matrix.

Last edited: Sep 18, 2016
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5. ### arfa branecall me arfValued Senior Member

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Not sure what you mean by "encapsulated" so much, but to be a tensor, a multidimensional array has to have certain transformation properties. You need, I think, to start with vector spaces and these have to have co- and contravariant properties.

A bare 2 x 2 x 2 cubic array of "numbers" or just symbols like a, b, c . . . with the transformation I've defined so far, might not qualify as a tensor. I need to define a direction vector somewhere along with what happens to it when it gets transported around under the rotation I've defined.

7. ### rpennerFully WiredRegistered Senior Member

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What do you mean?

The cyclic permutation group of 4 vertices? The rotational group of a square with distinct vertices in a 2-D embedding space? The rotational group of a square with indistinguishable vertices in a 2-D embedding space? The rotational group of a square with distinct vertices in a 2-D embedding space? The rotational group of an oriented square with indistinguishable vertices in a 3-D embedding space? The rotational group of an unoriented square with indistinguishable vertices in a 3-D embedding space?

8. ### arfa branecall me arfValued Senior Member

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Hmm. The rotation group or subgroup of symmetries of a square, or any polygon in a Euclidean domain, is closed under composition, any composition of rotations is a rotation.
This isn't true for reflections in the plane because compositions of reflections are rotations, and transposing a 2 x 2 matrix is analogous to reflecting two vertices of a square about a diagonal line.

In a matrix, all the elements are distinguished with indices. The transpose operation with another 'reflection' through a horizontal or vertical line of symmetry gives a 'rotation'.

So if say,

$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Then if A is any 2 x 2 matrix, we have:

$(A^{\intercal}\sigma_x)^2 = \sigma_x A \sigma_x$
and $\sigma_x = \sigma_x^{-1}$​

Last edited: Sep 21, 2016
9. ### arfa branecall me arfValued Senior Member

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$(A^{\intercal}\sigma_x)^2 = \sigma_x A \sigma_x$
and $\sigma_x = \sigma_x^{-1}$

Screwed up the notation. It should be:
$(A^{\intercal}\sigma_x)^{\intercal}\sigma_x = \sigma_x A \sigma_x$​

Still thinking about how to "repeat" either expression so you get the identity permutation on both sides.

10. ### arfa branecall me arfValued Senior Member

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Well, there are the equalities since, from the above:

$(A^{\intercal} \sigma_x)^{\intercal} = \sigma_x A$, so
$(A^{\intercal} \sigma_x) = (\sigma_x A) ^{\intercal}$,

by a symmetry, we have:

$(\sigma_x A^{\intercal}) = (A \sigma_x )^{\intercal}$

But $\sigma_x A \sigma_x$ "twice", is
$\sigma_x (\sigma_x A \sigma_x ) \sigma_x = \sigma_x^2 A \sigma_x^2 = A$​

11. ### arfa branecall me arfValued Senior Member

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Just a thought about what $\sigma_x A \sigma_x$ means.

Mathematically the rows are exchanged (transposed) and the columns are exchanged. It doesn't matter which is "done first".
However, it should be clear that

$\sigma_x (A \sigma_x) = ( \sigma_x A) \sigma_x$
Which although the parentheses are redundant, shows there is a rule of composition.

12. ### arfa branecall me arfValued Senior Member

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With the matrix transpose, we have a symmetric kind of operation:

$(\sigma_x A \sigma_x)^{\intercal} = \sigma_x A^{\intercal} \sigma_x$
But when $\sigma_x$ appears just once, either on the left or the right of A, the matrix transpose also transposes the position of $\sigma_x$:

$(\sigma_x A)^{\intercal} = A^{\intercal}\sigma_x,\; (A\sigma_x)^{\intercal} = \sigma_x A^{\intercal}$​