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Lets suppose matrices

$$ \begin{bmatrix} a & b\\ c & d\\ \end{bmatrix} $$ and $$ \begin{bmatrix} e & f\\ g & h\\ \end{bmatrix} $$ Assuming the first matrix is $C$ and the second one is $D$, replacing them in $CD=-DC$ we arrive at the following set of equations:

$ea+bg=-ea-fc$

$ec+dg=-ga-hc$

$af+bh=-eb-fd$

$cf+dh=-gb-hd$

Is there a way to find general solutions to the set of equations?

Math Lover
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David Sousa
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    Have a look at the Pauli matrices. – md2perpe Sep 21 '18 at 20:42
  • We're still in the beginning of the class, so we haven't learned that far. @md2perpe – David Sousa Sep 21 '18 at 20:52
  • But there you can find matrices with the wanted property. For example, take $C=\sigma_x=\begin{pmatrix}0&1\1&0\end{pmatrix}$ and $D=\sigma_z=\begin{pmatrix}1&0\0&-1\end{pmatrix}$. – md2perpe Sep 21 '18 at 20:57
  • yeah, but i wanted to follow my reasoning with the equations i provided. I was trying to see if there could be a way of solving that system of linear equations. @md2perpe – David Sousa Sep 21 '18 at 21:00
  • It's not a system of linear equations, you have variables multiplied by each other like $ea$, $af$ and so on. These are not linear equations. And for sure it's not easy to find a general solution to this system. – Mark Sep 21 '18 at 21:03
  • yeah you're right, so only by trial and error can this be solved ? @Mark – David Sousa Sep 21 '18 at 21:05
  • It's not really trial and error. There is logic here. I'll write it as an answer in a few minutes, a bit too long for a comment. – Mark Sep 21 '18 at 21:09
  • ok, thank you. @Mark – David Sousa Sep 21 '18 at 21:10

2 Answers2

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The best way is to use the so called elementary matrices. Let's say you have a matrix $A$, to make it easier let's assume it is really of size 2x2 like you need. It is a known fact that multiplying its first row by a scalar $\lambda$ is equivalent to multiplying $A$ by the matrix $P=\begin{pmatrix}\lambda&0\\0&1\end{pmatrix}$ on the left side. (which means $PA$ is the product that you get). In the same way multiplying the first column of $A$ by $\lambda$ is equivalent to multiplying $A$ by the same matrix $P$ on the right side. (you get the matrix $AP$).

So if you know that then it's easy to see that in your exercise you should take $\lambda=-1$, the matrix $D=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ and then you only have to find the matrix $C$. Take $C=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. And we really get:

$DC=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

$CD=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=-DC$

So as you can see the theory works. $DC$ is $C$ with the first row multiplied by $-1$, $CD$ is $C$ with the first column multiplied by $-1$. That's why it is very useful to know the properties of elementary matrices. Of course I still had to guess the matrix $C$ but it was much easier than if I had to guess both matrices from the beginning.

Mark
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Let $X$ be a $2$-by-$2$ matrix over a field $K$. For the sake of simplicity, we assume that $X$ is diagonalizable with eigenvalues $a$ and $b$. The $K$-linear map $T_X:V\to V$, where $V:=\text{Mat}_{2\times 2}(K)$, defined by $$T_X(Y):=XY+YX\text{ for all }Y\in V$$ is diagonalizable with eigenvalues $2a,2b,a+b,a+b$ (see this thread).

In the case that $\text{char}(K)\neq 2$, there exists a nonzero matrix $Y\in V$ such that $T_X(Y)=0$ if and only if $a=0$, $b=0$, or $a+b=0$. From the same thread, you can see that in case $a=0$ or $b=0$, every $Y\in\ker(T_X)$ satisfies $XY=YX=0$. Thus, there exists $Y\in V$ for which $T_X(Y)=0$ but $XY$ and $YX$ are both nonzero if and only if $a$ and $b$ are nonzero elements of $K$ such that $a+b=0$. This is why Mark's example above has to take a matrix of the form $\begin{bmatrix}+1&0\\0&-1\end{bmatrix}$.

In fact, if $K=\mathbb{R}$ (or any field of characteristic not equal to $2$), then we may without loss of generality assume that $X=\begin{bmatrix}+1&0\\0&-1\end{bmatrix}$. Then, $\ker(T_X)$ is spanned by $$\begin{bmatrix}0&1\\0&0\end{bmatrix}\text{ and }\begin{bmatrix}0&0\\1&0\end{bmatrix}\,.$$ That is, any matrix $Y$ of the form $$Y=\begin{bmatrix}0&p\\q&0\end{bmatrix}\text{ with }(p,q)\in\mathbb{R}^2\setminus\big\{(0,0)\big\}$$ satisfies $XY+YX=T_X(Y)=0$ but $XY$ and $YX$ are nonzero.

In the case $K=\mathbb{C}$, we can see why the Pauli matrices have the form $$\sigma_x=\begin{bmatrix}0&1\\1&0\end{bmatrix}\,,\,\,\sigma_y=\begin{bmatrix}0&-\text{i}\\+\text{i}&0\end{bmatrix}\,,\text{ and }\sigma_z=\begin{bmatrix}+1&0\\0&-1\end{bmatrix}\,.$$ If we start with setting $\sigma_z$ to be $\begin{bmatrix}+1&0\\0&-1\end{bmatrix}$, then $\ker(\sigma_z)$ consists of matrices of the form $$\mu(p,q):=\begin{bmatrix}0&p\\q&0\end{bmatrix}\text{ with }p,q\in\mathbb{C}\,.$$

If we want the matrix $\mu(p,q)$ to also have eigenvalues $+1$ and $-1$, then $pq=1$ must hold. We can then take $\sigma_x$ in the form $\mu(1,1)=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, and look at $\ker(\sigma_x)\cap \ker(\sigma_z)$ to see whether it is nontrivial.

Since $\ker(\sigma_x)$ is spanned by $$\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\text{ and }\begin{bmatrix}1&-1\\1&-1\end{bmatrix}\,,$$ $\ker(\sigma_x)\cap\ker(\sigma_z)$ consists of matrices of the form $$\nu(r):=\begin{bmatrix}0&-r\\+r&0\end{bmatrix}\,,\text{ where }r\in\mathbb{C}\,.$$ Thus, a matrix $\nu(r)$ has eigenvalues $+1$ and $-1$ if and only if $r^2=-1$, or equivalently, $r\in\{-\text{i},+\text{i}\}$. With the choice $r=+\text{i}$, we obtain the matrix $$\sigma_y=\begin{bmatrix}0&-\text{i}\\+\text{i}&0\end{bmatrix}\,.$$

Batominovski
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