Show that $XY=0$ or $YX=0$ We have $X,Y$ $(2×2)$ matrices with complex entries and $X=A^{2}-B^{2}$ and $Y=AB-BA$. We know that $\det(X)=\det(Y)=0$. Show that $XY=0$ or $YX=0$.
I see that Trace of $Y$ is $0$ and $\det(Y)$ is also $0$ so by $C-H$, $Y^{2}=0$. I also saw how we can write $(X+Y)$ as $(A-B)(A+B)$ but nothing more.
 A: If $X=0$ or $Y=0$, there is nothing to prove. Assume that $X,Y\ne0$. Recall the following identity for $2\times2$ matrices:
$$
\det(X+Y)+\det(X-Y)=2(\det(X)+\det(Y)).
$$
By assumption, $\det(X)=\det(Y)=0$. Therefore
$$
\det(X+Y)+\det(X-Y)=0.\tag{1}
$$
Let $D=A-B$ and $S=A+B$. Then $DS=X+Y$ and $SD=X-Y$. Hence $(1)$ implies that $\det(SD)+\det(DS)=0$, i.e., $2\det(D)\det(S)=0$. Therefore one of $S$ or $D$ must be singular.
It follows that $Av=\pm Bv$ for some nonzero vector $v$. Thus
$$
Xv=(A^2-B^2)v=AAv-BBv=A(\pm Bv)-B(\pm A)v=\mp Yv.\tag{2}
$$
Since $X$ and $Y$ are singular, nonzero and $2\times2$, they must be rank-one matrices. Let $X=x_1x_2^\ast$ and $Y=y_1y_2^\ast$ for some nonzero vectors $x_1,x_2,y_1$ and $y_2$. Note that $y_1\perp y_2$ because
$$
y_2^\ast y_1=\operatorname{tr}(Y)=\operatorname{tr}(AB-BA)=0.
$$
Now return to $(2)$. There are two possibilities:

*

*$Xv=\mp Yv=0$. Then $x_1(x_2^\ast v)=\mp y_1(y_2^\ast v)=0$. Therefore both $x_2$ and $y_2$ are orthogonal to $v$. Hence $x_2$ is parallel to $y_2$. In turn, $x_2$ is orthogonal to $y_1$ and $XY=x_1(x_2^\ast y_1)y_2^\ast=0$.

*$Xv=\mp Yv\ne0$. Then $x_1(x_2^\ast v)=\mp y_1(y_2^\ast v)\ne0$. Therefore $x_1$ is parallel to $y_1$. In turn, $x_1$ is orthogonal to $y_2$ and $YX=y_1(y_2^\ast x_1)x_2=0$.

