If $A,B \in \mathcal{M_2}(\mathbb{R})$ and $A^2+B^2=AB$, does it follow that $A$ and $B$ commute?

Solution 1:

Let $w$ be a primitive cubic root of unity and $X=A+wB$. If $A^2+B^2=AB$, then $$ X\overline{X}=(A+wB)(A+w^2B)=wBA+(1+w^2)AB=w(BA-AB). $$ Therefore $w^2\det(BA-AB)=|\det(X)|^2$ is real and $\det(BA-AB)$ must be zero. It follows that $X$ is singular. As it is also $2\times2$, we may write $X=uv^\ast$ for some $u,v\in\mathbb C^2$. Hence $$ |v^\top u|^2=\operatorname{tr}\left(u(\overline{v^\top u})v^\top\right) =\operatorname{tr}(X\overline{X})=w\operatorname{tr}(AB-BA)=0 $$ and $v^\top u=0$. Consequently, $w(AB-BA) = X\overline{X} = u(\overline{v^\top u})v^\top = 0$, i.e. $AB=BA$.

Solution 2:

$\textbf{Proposition 1}$. Let $A,B\in M_2(\mathbb{C})$ s.t. $A^2+B^2=AB$. Then $A,B$ have a common eigenvector.

$\textbf{Proof}$. Let $A=X+Y,B=-jX-j^2Y$ where $j=\exp(2i\pi/3)$. Then

$XY=j^2YX, AB-BA=(j-j^2)(XY-YX)$.

Note that $\det(XY)=j\det(YX)$ implies that $\det(XY)=0$. Clearly $\det(AB-BA)=0$ and $rank(AB-BA)\leq 1$. That implies that $A,B$ have a common eigenvector. $\square$

$\textbf{Proposition 2}$. Let $A,B\in M_2(\mathbb{R})$ s.t. $A^2+B^2=AB$. Then $AB=BA$.

$\textbf{Proof}$. Let $u\in\mathbb{C}^2\setminus\{0\}$ s.t. $Au=\lambda u,Bu=\mu u$.

1. If we cannot choose $u$ in $\mathbb{R}^2\setminus\{0\}$ then necessarily $\lambda,\mu\notin\mathbb{R}$ and $A\overline{u}=\overline{\lambda}\overline{u},B\overline{u}=\overline{\mu}\overline{u}$. Since $\{u,\overline{u}\}$ is a basis of $\mathbb{C}^2$, we deduce that $AB=BA$.

2. If we can choose $u$ in $\mathbb{R}^2\setminus\{0\}$ then necessarily $\lambda,\mu\in\mathbb{R}$ and the eigenvalues of $A,B$ are all real. Thus, we may assume that $A,B$ are upper-triangular real matrices in the form

$A=\begin{pmatrix} a&b\\0&c\end{pmatrix},B=\begin{pmatrix}d&e\\0&f\end{pmatrix}$ where $f\in\{-jc,-j^2c\},d\in\{-ja,-j^2a\},b.tr(A)+e.tr(B)=ae+bf$.

We conclude that $a=c=d=f=0$ and $AB=BA$. $\square$

Note that there are solutions over $\mathbb{C}$ that do not commute.

Solution 3:

Proposition: Let $A,B \in \mathcal{M_2}(\mathbb{R})$ such that $A^2+B^2=AB$. Then $AB=BA$.

Disclaimer: This proof is absolutely awful.

Proof. Note that for every real number $\lambda\in\Bbb{R}$ we have $$(\lambda A)^2+(\lambda B)^2=\lambda^2(A^2+B^2)=\lambda^2(AB)=(\lambda A)(\lambda B).$$

If $\det A\neq0$ then after an appropriate change of basis we have $$A=\begin{pmatrix}1&0\\0&\lambda\end{pmatrix} \qquad\text{ or }\qquad A=\begin{pmatrix}1&\lambda\\0&1\end{pmatrix} \qquad\text{ or }\qquad A=\begin{pmatrix}\lambda&-1\\1&\lambda\end{pmatrix},$$ for some $\lambda\in\Bbb{R}$. Let $B=\tbinom{a\ b}{c\ d}$. We treat the three cases separately:

1. Plugging $A=\tbinom{1\ 0}{0\ \lambda}$ and $B$ into the equation yields the system of equations \begin{eqnarray*} 1+a^2+bc&=&a,\\ b(a+d)&=&b,\\ c(a+d)&=&\lambda c,\\ \lambda^2+d^2+bc&=&\lambda d. \end{eqnarray*} If $bc=0$ then $a^2-a+1=0$ which is impossible. Then $\lambda=a+d=1$ and so $A$ is the identity matrix, so certainly $AB=BA$.

2. Plugging $A=\tbinom{1\ \lambda}{0\ 1}$ and $B$ into the equation yields the system of equations \begin{eqnarray*} 1+a^2+bc&=&a+\lambda c,\\ 2\lambda+b(a+d)&=&b+\lambda d,\\ c(a+d)&=& c,\\ 1+d^2+bc&=&d. \end{eqnarray*} If $bc=0$ then $d^2-d+1=0$ which is imposible. It follows that $a+d=1$ and hence that $2\lambda=\lambda d$. If $\lambda=0$ then $A$ is the identity matrix so certainly $AB=BA$. Otherwise $d=2$ and hence $bc=-3$ and $a=-1$, which leads to $\lambda c=0$, a contradiction.

3. Plugging $A=\tbinom{\lambda\ -1}{1\ \hphantom{-}\lambda}$ and $B$ into the equation yields the system of equations \begin{eqnarray*} \lambda^2-1+a^2+bc&=&\lambda a-c,\\ -2\lambda+b(a+d)&=&\lambda b-d,\\ 2\lambda+c(a+d)&=&a+\lambda c,\\ \lambda^2-1+d^2+bc&=&b+\lambda d. \end{eqnarray*} Adding the second and third, and subtracting the last from the first, yields $$(b+c)(a+d)=\lambda(b+c)+(a-d) \qquad\text{ and }\qquad a^2-d^2=\lambda(a-d)-(b+c).$$ Isolating $a-d$ from the former and $b+c$ from the latter and plugging it in shows that $$a-d=(b+c)(a+d-\lambda)=(a-d)(\lambda-a-d)(a+d-\lambda)=-(a+d-\lambda)^2(a-d),$$ which shows that $a=d$. It follows that $b=-c$ and we are left with the system \begin{eqnarray*} \lambda^2-1+a^2-b^2&=&\lambda a-b,\\ -2\lambda+2ab&=&\lambda b-a. \end{eqnarray*} The latter shows that $a(2b+1)=\lambda(b+2)$, and multiplying the former by $(b+2)^2$ and substituting and cleaning up yields $$3a^2(b^2-b+1)=(b+2)^2(b^2-b+1),$$ where $b^2-b+1\neq0$ because $b\in\Bbb{R}$. It follows that $b=-2\pm\sqrt{3}a$ and $\lambda=2a\mp\sqrt{3}$ correspondingly. Then $$A=\begin{pmatrix} 2a\mp\sqrt{3}&-1\\ 1&2a\mp\sqrt{3} \end{pmatrix} \qquad\text{ and }\qquad B=\begin{pmatrix} a&-2\pm\sqrt{3}{a}\\ 2\mp\sqrt{3}{a}&a \end{pmatrix},$$ and a routine check verifies that again $AB=BA$.

Finally, if $\det A=0$ then after an appropriate change of basis we have $$A=\begin{pmatrix}\lambda &0\\0&0\end{pmatrix} \qquad\text{ or }\qquad A=\begin{pmatrix}0&\lambda\\0&0\end{pmatrix},$$ for some $\lambda\in\Bbb{R}$, where the proposition is trivial if $\lambda=0$. If $\lambda\neq0$ a routine check as before shows that for the first form there is no corresponding matrix $B$ satisfying the identity, and for the second form we see that $B$ must be of the form $$B=\begin{pmatrix}0&\mu\\0&0\end{pmatrix},$$ for some $\mu\in\Bbb{R}$, which shows that $AB=BA$.