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The question is:

Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?

From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?

Edit: $$A^2+B^2=AB\implies\\ A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\ A^3+(A^2+B^2)B=A^2B \implies \\ A^3+A^2B+B^3=A^2B \implies \\ A^3+B^3=0 $$

Ben Grossmann
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user35603
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    No. For example, $A = 0$, $B = \pmatrix{0&1\0&0}$ satisfies $A^3 + B^3 = 0 + 0 = 0$ but the only polynomial in $A$ is $0$. – Travis Willse May 02 '16 at 19:32
  • I'm not sure how much it helps, but FWIW, $0=(A+B)(A^2-AB+B^2)+(A^2-AB+B^2)(A+B)=(BA-AB)A+B(BA-AB)$ (using also $A^3+B^3=0$) – m7e May 02 '16 at 21:48

2 Answers2

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The answer is yes. We shall construct a concrete example below. First, suppose $A$ and $B$ are invertible and let $C=AB^{-1}$. From $A^2+B^2=AB$ we get $C + A^{-1}C^{-1}A = I$ and $A^{-1}CA=(I-C)^{-1}$. Therefore the spectrum of $C$ is invariant under the transformation $z\mapsto \frac1{1-z}$.

Let $w$ be the cubic root of unity. If $A^2+B^2=AB$, then $(A+wB)(A+w^2B)=w(BA-AB)$. So, if we want $BA-AB$ to be invertible, neither $-w$ nor $-w^2$ can be an eigenvalue of $C$. As $-w$ and $-w^2$ are fixed points of $z\mapsto\frac1{1-z}$, the spectrum of $C$ has to be splittable into three cycles of the form $\{z,\ \frac1{1-z},\ \ 1-\frac1z\}$, where $z\ne-w,-w^2$, and for every feasible pair of invertible matrices $A,B$, their dimensions must be divisible by 3.

Now we can construct an example on $M_3(\mathbb C)$. Take $z=2$ and $C=\operatorname{diag}(2, -1, \frac12)$. So $(I-C)^{-1}=\operatorname{diag}(-1, \frac12, 2)$. To make $A^{-1}CA=(I-C)^{-1}$ or $A^2+B^2=AB$, we pick $A=\pmatrix{0&0&1\\ 1&0&0\\ 0&1&0}$ and $B=C^{-1}A=\pmatrix{0&0&\frac12\\ -1&0&0\\ 0&2&0}$. One can readily verify that $BA-AB=\pmatrix{0&-\frac32&0\\ 0&0&-\frac32\\ 3&0&0}$ is invertible.

user1551
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  • You have a solution for $n=3$. Is it possible to have solutions for all $n=3k$ ? See https://math.stackexchange.com/q/3109173 – Jean Marie Dec 30 '19 at 21:53
  • @JeanMarie Yes. Simply repeat each of $A$ and $B$ $k$ times in a block-diagonal construction. – user1551 Dec 30 '19 at 22:20
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We consider the equations
(1) $A^2+B^2=AB$,
(2) $A^2+B^2=2AB$.
A couple $(A,B)$, solution of (2), is simultaneously triangularizable, and, moreover, when $n=2$, $AB=BA$.

Eq. (1) has not the second property: indeed $$A=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},\ B=\begin{pmatrix}\dfrac{1-i\sqrt{3}}{2}&0\\\dfrac{3+i\sqrt{3}}{2}&-1\end{pmatrix}$$ is a solution s.t. $AB\not= BA$. Note that $\mathrm{spectrum}(A)=\{e^{\pm 2i\pi/3}\}$ and $A^3=I_2$, $B^3=-I_2$; moreover $(AB-BA)^2=0$. I think that Eq. (1) has the first property, or at least, $AB-BA$ is nilpotent, but I don't know how to prove it. We see, in the following, that, a solution, in general position, satisfies $AB=BA$.

Proposition. We consider the equation in the unknown $X$: $X^2-AX+A^2=0$. Let $\mathrm{spectrum}(A)=(\lambda_i)_i$ denote the complete spectrum of $A$. If, for every $i\not= j$ $\lambda_i^3\not=\lambda_j^3$, then any solution $X$ satisfies $AX=XA$.

Proof. Since $A$ is diagonalizable we may assume $A=\mathrm{diag}(\lambda_i)$. Thus $A^3=\mathrm{diag}(\lambda_i^3)$ has distinct eigenvalues; since $X$ and $A^3$ commute, $X$ is a diagonal matrix and we are done.

EDIT. There are solutions of $A^3+B^3=0$ s.t. $AB-BA$ is invertible; for example, choose $A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\ B=\begin{pmatrix}0&0&0\\1&0&0\\0&-1&0\end{pmatrix}$; of course $A^2+B^2\not= AB$.

user26857
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  • Thanks for response. It also interesting if $B^3=-A^3$ $()$, can we say then that $B$ is a function of $A$ ($B=f(A)$). One solution of $()$ is always exists $B=-A$. – user35603 May 05 '16 at 15:34