# Abstract

Under matrix multiplication, different shaped matrices (diagonal, upper/lower triangular and anti diagonal variants etc.) form various other shapes as a result. For example in LU decomposition, a lower and upper matrix form a square matrix. I attempt to search for a group and explore this concept a little.

# Introduction

We can draw a Cayley table under the matrix multiplication, however this is a non-commutative operation so we do not expect the table to be symmetric.

Let us define, $$s$$ as a square positive matrix (all elements greater than zero). $$d$$ and $$a$$ as diagonal and anti diagonal positive matrices respectively, then $$nw,ne,sw$$ and $$se$$ (compass directions) as triangular positive matrices around the respective corner.

//(Examples for explicit)

$\begin{array}{ c| c c c c c c c} & s & sw & ne & se & nw & d & a \\ \hline s & s & s & s & s & s & s & s \\ sw & s & sw & s & se & s & sw & se \\ ne & s & s & ne & s & nw & ne & nw \\ se & s & s & se & s & sw & se & sw \\ nw & s & nw & s & ne & s & nw & ne \\ d & s & sw & ne & se & nw & d & a \\ a & s & nw & se & ne & sw & a & d \\ \end{array}$

It is clear from this table that the element $$s$$ acts like $$0$$ as when it is multiplied to anything it creates itself. Then the element $$d$$ acts like $$1$$ as when it is multiplied to anything it leaves it the same, (identity). In some sense $$a$$ is then like $$-1$$ as it switches some property of a given element, a left multiplication of $$a$$ changes $$sX$$ to $$nX$$ and $$nX$$ to $$sX$$, where $$X$$ is $$e$$ or $$w$$. a right multiplication of $$a$$ changes $$Xw$$ to $$Xe$$ and $$Xe$$ to $$Xw$$, where $$X$$ is $$n$$ or $$s$$.

With this in mind a left and right multiplication of some element with $$a$$ will change both parts, for example, $$(a)(se)(a)=(nw)$$.

There cannot be an inverse for every operation when we insist that the matrices are positive, thus they cannot form a group.

# New

Exploit symmetries, ${\begin{bmatrix} a & b \\ c & d \end{bmatrix}} \to \Bigg[ \begin{matrix} b & \\ a & d \\ c & \end{matrix}$

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix} \to \Bigg[ \begin{matrix} af+bh & \\ ae+bg & cf+dh \\ ce+dg & \end{matrix}$

$\Bigg[ \begin{matrix} b & \\ a & d \\ c & \end{matrix}^T = \Bigg[\begin{matrix} c & \\ a & d \\ b & \end{matrix}$

# Complex

With complex numbers $$(a+ib),(c+id)$$ could make $(a+ib),(c+id) \to \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

Thus with an additional pair $$(e+if),(g+ih)$$ we have $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae-bf & af+be \\ cg-dh & ch+dg \end{bmatrix}$

Which represents a term for term product of two complex vectors, however carries the same construction as two by two matrix multiplication! One could ask the question, do there exist pairs of matrices which give the same written equation when the dot is interpreted as matrix multiplication or as the complex vector system? A trivial solution is all elements are zero, so at least one.

A non trivial solution would would be the set of equations $ae-bf = ae+bg \\ af+be = af+bh \\ cg-dh = ce+dg \\ ch+dg = cf+dh$

Which reduce to $-f = g \\ e = h \\ cg-dh = ce+dg \\ ch+dg = cf+dh$

However if $$c=0$$ (the left hand matrix is upper triangular) we would have that $$-h = g , g = h$$, which is only satisfied by $$g=h=0$$... Therefore from the top two equations, also $$f=e=0$$, then $$a,b$$ and $$d$$ can take any value as the equations are independent of these values.

The operaration here is uncoupled so could make a strange change such as defining a new product which is $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \# \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae-dh & af+dg \\ cg-bf & ch+be \end{bmatrix}$

Term by term deliberatly picking 0 and 1${\begin{bmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \end{bmatrix}}\#{\begin{bmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \end{bmatrix}} = {\begin{bmatrix} a_{00}b_{00}-a_{11}b_{11} & a_{00}b_{01}+a_{11}b_{10} \\ a_{10}b_{10}-a_{01}b_{01} & a_{10}b_{11}+a_{01}b_{00} \end{bmatrix}}$

After a fair amount of thinking this has a general formula $a\#b=M_{ij}=\sum_{k=0}^1 (-1)^{\delta^k_1\delta^j_0}a_{i \oplus k,k}b_{i \oplus k,j \oplus k}$ where $$\oplus$$ is the logical XOR operation.

We have the following results $\begin{bmatrix}1&0\\0&0\end{bmatrix}\#{\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}=\begin{bmatrix}1&1\\0&0\end{bmatrix}\\ \begin{bmatrix}0&1\\0&0\end{bmatrix}\#{\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}=\begin{bmatrix}0&0\\-1&1\end{bmatrix}\\ \begin{bmatrix}0&0\\1&0\end{bmatrix}\#{\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}=\begin{bmatrix}0&0\\1&1\end{bmatrix}\\ \begin{bmatrix}0&0\\0&1\end{bmatrix}\#{\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}=\begin{bmatrix}-1&1\\0&0\end{bmatrix}\\$ The relationship does not commute ${\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}\#\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I\\ {\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}\#\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}=i\sigma_y\\ {\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}\#\begin{bmatrix}0&0\\1&0\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}=\sigma_x\\ {\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}}\#\begin{bmatrix}0&0\\0&1\end{bmatrix}=\begin{bmatrix}-1&0\\0&1\end{bmatrix}=-\sigma_z\\$

Which are an interesting basis.

The transform appears to follow the relationship $1\#\begin{bmatrix}a&b\\c&d\end{bmatrix}=1\#\begin{bmatrix}a&0\\0&0\end{bmatrix}+1\#\begin{bmatrix}0&b\\0&0\end{bmatrix}+1\#\begin{bmatrix}0&0\\c&0\end{bmatrix}+1\#\begin{bmatrix}0&0\\0&d\end{bmatrix}$

also

$\begin{bmatrix}1&1\\1&1\end{bmatrix}\#\begin{bmatrix}a&0\\0&0\end{bmatrix}=\begin{bmatrix}a&a\\a&a\end{bmatrix}\#\begin{bmatrix}1&0\\0&0\end{bmatrix}=a(\begin{bmatrix}1&1\\1&1\end{bmatrix}\#\begin{bmatrix}1&0\\0&0\end{bmatrix})$

The operation has a left identity $\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}\#\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}$

However it is not possible to define a right identity! Using the same matrix above will keep the elements the same bu swap those in the right hand column. Most likely a strange artifact of the way we came to the transform.

We can state however that $\bigg( {\begin{bmatrix} a & b \\ c & d \end{bmatrix}} \# {\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}}\bigg) \# {\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}} = {\begin{bmatrix} a & b \\ c & d \end{bmatrix}}$

This is strange! as ${\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}} \# {\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}} = {\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}}$

So the operation cannot be associative! For some element $$\iota \equiv 1$$ we then have that $$\iota a = (a \iota) \iota \ne a\iota$$.

We can define a left inverse, for some $$A\#B=C$$ such that $$(A^{-1}\#A)\#B=A^{-1}\#C=B$$. For such a system we solve for the $$a,b,c,d$$ such that $\begin{bmatrix} e & 0 & 0 &-h \\ f & 0 & 0 & g \\ 0 & -f & g & 0 \\ 0 & e & h & 0 \end{bmatrix}\begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$ Which results in $\begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix}=\frac{1}{eg+fh}\begin{bmatrix}g&h&0&0\\0&0&-h&g\\0&0&e&f\\-f&e&0&0\end{bmatrix} \begin{bmatrix}1 \\ 0 \\ 1 \\ 0 \end{bmatrix}=\frac{1}{eg+fh}\begin{bmatrix}g\\-h\\e\\-f\end{bmatrix}$

This gives the general formula for the $$\#$$ inverse of a 2 by 2 matrix as $\begin{bmatrix} a&b\\c&d\end{bmatrix}^{-1}=\frac{1}{ac+bd}\begin{bmatrix}c & -d \\ a & -b \end{bmatrix}$

This is fondly reminiscent of the steps taken to invert a regular matrix under matrix multiplication, exept exchanges are of rows rather than on diagonal terms and two negatives in a different place. Also there will still exist $$\#$$ singular matrices when $$ac=-bd$$. This would make such a quantity the $$#$$ determinant of the matrix...

Interesting question, if a matrix is normally singular $$ad-bc=0$$, can it also be $$\#$$ singular, $$ac+bd=0$$? There can be if $$a$$ and $$b$$ are both $$0$$, or for certain combinations otherwise, some with complex entries...

A desirable scenario would now be a regular equation of the form $Ax=b \\ \Upsilon\# Ax=\Upsilon\# b$

If we can use an $$\Upsilon$$ that is the $$\#$$ inverse of $$A$$, then we know $x=\Upsilon\# b$

However it is not yet clear how to define the $$\#$$ operation to a vector. This can be explored, we must find the general expression for a matrix to matrix $$\#$$ operation. In the mean time we can set up a scenario we know the solution to, attempting to reverse engineer the result! $\begin{bmatrix}n&0\\0&n\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}nx_1\\nx_2\end{bmatrix}$

We know the $$\#$$ inverse of the identity matrix,

$\frac{1}{n}\begin{bmatrix}1&0\\1&0\end{bmatrix}\#\begin{bmatrix}n&0\\0&n\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\frac{1}{n}\begin{bmatrix}1&0\\1&0\end{bmatrix}\#\begin{bmatrix}nx_1\\nx_2\end{bmatrix}$

The factors of $$n$$ will cancel, however this reveals to us that the left identity is still an identity for a vector. This is fairly obvious, but now known for sure. A less easy example would be

$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}ax_1+bx_2\\cx_1+dx_2\end{bmatrix}$

We can use our $$\#$$ inverse already found to give $\bigg(\frac{1}{ac+bd}\begin{bmatrix}c & -d \\ a & -b \end{bmatrix}\#\begin{bmatrix}a&b\\c&d\end{bmatrix}\bigg)\begin{bmatrix}x_1\\x_2\end{bmatrix}=\frac{1}{ac+bd}\begin{bmatrix}c & -d \\ a & -b \end{bmatrix}\#\begin{bmatrix}ax_1+bx_2\\cx_1+dx_2\end{bmatrix}\\ {\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\frac{1}{ac+bd}\begin{bmatrix}c & -d \\ a & -b \end{bmatrix}\#\begin{bmatrix}ax_1+bx_2\\cx_1+dx_2\end{bmatrix} \\ \begin{bmatrix}x_1\\x_1\end{bmatrix}=\frac{1}{ac+bd}\begin{bmatrix}c & -d \\ a & -b \end{bmatrix}\#\begin{bmatrix}ax_1+bx_2\\cx_1+dx_2\end{bmatrix} \\$

We can define the concept of a “cross inverse” such that for any \(A.B=C\