If c and d are any two 2 2 matrices such that c d d c c c is not a zero matrix what is matrix d

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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 $$\sigma_y=\begin{bmatrix}0&-\text{i}\\+\text{i}&0\end{bmatrix}\,.$$