# If $(A-B)^2=AB$, prove that $\det(AB-BA)=0$.

Let $A,B\in M_{n}(\mathbb{Q})$. If $(A-B)^2=AB$, prove that $\det(AB-BA)=0$.

I considered the function $f:\mathbb{Q}\rightarrow \mathbb{Q}$, $f(x)=\det(A^2+B^2-BA-xAB)$ and I obtained that: $$f(0)=\det(A^2+B^2-BA)=\det(2AB)=2^n\det(AB)$$ $$f(1)=\det(A^2+B^2-BA-AB)=\det((A-B)^2)=\det(AB)$$ $$f(2)=\det(A^2+B^2-BA-2AB)=\det((A-B)^2-AB)=\det(AB-AB)=0$$

I don't have any other idea.

• You can easily find such matrices over $\mathbb{R}$, so it is plausible such rational matrices exist as well. For example $A = \dfrac{3+\sqrt{5}}{2}I$ and $B=I.$ Apr 18, 2017 at 23:45
• I wonder if there is an example of such $A$ and $B$, even with complex entries, such that $AB-BA\neq 0$. Apr 19, 2017 at 3:09
• @JonasMeyer I found one, see my edited answer. The entries are real numbers though. Apr 21, 2017 at 1:07

The idea is to use $$(-3\pm\sqrt 5)/2$$. We begin with

\begin{align} (A+xB)(A+x'B) &= A^2 + x BA + x' AB + xx' B^2 \\ &= A^2 + (x' + x) AB+x(BA-AB)+xx'B^2.\end{align}

Now we find numbers $$x$$, $$x'$$ such that $$x'+x = -3$$, $$xx'=1$$. These numbers are the roots of the quadratic equation $$\lambda^2 +3\lambda +1 = 0.$$ Thus, we have $$x = \frac{-3+\sqrt 5}2$$ and $$x'= \frac{-3-\sqrt 5}2$$.

With these, we have \begin{align} (A+xB)(A+x'B) &= A^2 -3AB + B^2 + x(BA-AB) \\ &= BA-AB + x(BA-AB) = (BA-AB)(1+x).\end{align} We now have that $$\det(A+xB)(A+x'B) =q \in \mathbb{Q},$$ and $$q=(1+x)^n \det(BA-AB).$$ Then it follows by $$A, B\in M_n(\mathbb{Q})$$ and $$(1+x)^n\in\mathbb{R} \backslash \mathbb{Q}$$ that $$\det(BA-AB)=0$$.

An example with $$AB\neq BA$$

For the example that @Jonas Meyer requested, here it is: $$A=\begin{pmatrix} 1 & 1 \\ 0 & \frac{-3+\sqrt 5}2\end{pmatrix}, \ \ B=\begin{pmatrix} \frac{3+\sqrt 5}2 & 1 \\ 0 & -1\end{pmatrix}.$$ Then $$A-B= \begin{pmatrix} \frac{-1-\sqrt 5}2 & 0 \\ 0 & \frac{-1+\sqrt 5}2\end{pmatrix},$$ $$(A-B)^2 = \begin{pmatrix} \frac{3+\sqrt 5}2 & 0 \\ 0 & \frac{3-\sqrt 5}2\end{pmatrix} = AB.$$ But, $$BA =\begin{pmatrix} \frac{3+\sqrt 5}2 & \sqrt 5 \\ 0 & \frac{3-\sqrt 5}2\end{pmatrix} \neq AB.$$

• Could you clarify how you conclude that $q\in\mathbb Q$? Apr 19, 2017 at 2:23
• Its conjugate inside $\mathbb{Q}(\sqrt 5)$ is the same with itself. Apr 19, 2017 at 2:25
• Ah right, thanks. So is there a counterexample over $\mathbb C$ then? Apr 19, 2017 at 2:26
• Not sure right now, but some modification of Arin's example would give a counterexample. Apr 19, 2017 at 2:30
• Can you elaborate on how to see that $q$ is fixed under conjugation in $\mathbb{Q}(\sqrt{5})$? Apr 19, 2017 at 2:32

The statement is true without the assumption that $A,B$ have rational entries. As in i707107's answer, let $$x=\frac{-3-\sqrt{5}}2,\,x'=\frac{-3+\sqrt{5}}2$$ so that $x+x'=-3$, $xx'=1$. Let $X=A+xB$, $Y=A+x'B$. Then $$\begin{eqnarray*} (1+x')XY-(1+x)YX &=&(x'-x)(A^2+xx'B^2)+(x'(1+x')-x(1+x))AB\\ &&{}+(x(1+x')-x'(1+x))BA\\ &=&(x'-x)\left[A^2+B^2+(1+x'+x)AB-BA\right]\\ &=&(x'-x)\left[(A-B)^2-AB\right]\\ &=&0. \end{eqnarray*}$$ Thus $XY=kYX$ where $$k=\frac{1+x}{1+x'}$$ Note that $|k|>1$. Pick an eigenvector $v$ for $X$ whose eigenvalue $\lambda$ has maximal magnitude. Then $$X(Yv)=kYXv=(k\lambda)(Yv).$$ If $\lambda=0$ then $YXv=XYv=0$. If $\lambda\neq0$ then $|k\lambda|>|\lambda|$, so by assumption $k\lambda$ can't be an eigenvalue of $X$. This implies $Yv=0$, so again $XYv=0$ and $YXv=\lambda Yv=0$. In either case we have $(XY-YX)v=0$, so $XY-YX$ is singular. Finally since $$XY-YX=(x'-x)(AB-BA),$$ we conclude $AB-BA$ is singular.

To give a bit more insight, suppose we started with an arbitrary homogeneous degree 2 constraint on $A,B$: $$c_2A^2+c_1AB+c_1'BA+c_0B^2=0.$$ If we replace $A,B$ by commuting variables $a,b$, the corresponding polynomial would factor over $\mathbb C$: $$c_2a^2+(c_1+c_1')ab+c_0b^2=(\alpha a+\beta b)(\gamma a+\delta b).$$ Let $X=\alpha A+\beta B$ and $Y=\gamma A+\delta B$. If $A$ and $B$ commuted we'd have $XY=0$, but instead we get $$XY=(\alpha\delta-c_1)[A,B]$$ where $[A,B]=AB-BA$ is the Lie bracket. Note that $[X,Y]=(\alpha\delta-\beta\gamma)[A,B]$, so $$(\alpha\delta-\beta\gamma)XY=(\alpha\delta-c_1)[X,Y].$$ Unless a coefficient happens to vanish, this gives $XY=kYX$ for some $k$. When $k$ is not a root of unity this is quite a restrictive constraint (eg $XY$ must be singular).

• My first thought was that the result extends to (but may not beyond) $\Bbb{R}$, so this is interesting to see the answer carries over to all of $\Bbb{C}$. (+1) Apr 19, 2017 at 3:16
• @SangchulLee I think you could actually deduce the result over $\mathbb C$ from the result over $\mathbb Q$; extend to any finite extension $F$ of $\mathbb Q$ using the embedding $\rm{End}_{F}(F^n)\to\rm{End}_{\mathbb Q}(F^n)$, and then to the algebraic closure of $\mathbb Q$. Finally frame the problem as asking whether an algebraic variety is empty and use weak Hilbert's Nullstellensatz. Apr 19, 2017 at 6:18

In fact, there is a much stronger result.

Proposition. Let $$a,b,c,d\in\mathbb{C}$$. If $$A,B\in M_n(\mathbb{C})$$ satisfy $$aA^2+bB^2+cAB+dBA=0_n$$, then, generically (for example, randomly choose $$a,b,c,d$$), $$A,B$$ are simultaneously triangularizable (for short, ST).

Proof. We follow the last part of the stewbasic's good post. We put $$X=\alpha A+\beta B,Y=\gamma A+\delta B$$; since generically $$\alpha\delta\not= \beta\gamma$$, it suffices to show that $$X,Y$$ are ST.

We obtain (generically) $$XY=kYX$$ for some complex number $$k$$. Generically, $$k\not=1$$ and $$k$$ is not a primitive root of unity; then, by a result of Drazin, $$X,Y$$ are ST.

Remark 1. Of course, there are $$a,b,c,d$$ s.t. $$A,B$$ are not necessarily ST. For instance, consider $$AB+BA=0_2$$ with $$A=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\, B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$.

Remark 2. That is linked to the concept of quasi-commutative matrices. There are two non-equivalent definitions:

i) $$A,B$$ are quasi-commutative iff $$AB-BA$$ commute with $$A,B$$. By a result from McCoy, $$A,B$$ are always ST.

ii) $$A,B$$ are quasi-commutative iff $$AB=kBA$$ where $$k$$ is a complex number. This is the definition we are interested in.

• What says that result of Drazin? Jan 5, 2020 at 21:07
• Anyway, I think we also need $k\ne0$. Then, from $XY=kYX$ we get that $XY$ and $YX$ are nilpotent. Thus $p(X,Y)(XY-YX)$ is nilpotent (for $p$ a polynomial). From McCoy's theorem it follows that $X,Y$ are simultaneously triangularizable. Jan 5, 2020 at 21:52
• @user26857 , see math.wisc.edu/hans/paper_archive/scanned_papers/hs152.pdf
– user91684
Jan 5, 2020 at 22:41