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Let $A$ and $B$ be two $2 \times 2$ matrices with real entries such that $A^2+B^2=AB$ then prove that $(AB-BA)^2=O$, where $O$ is the zero matrix.

I tried with assuming two matrices but it becomes very lengthy and difficult to calculate. Then I tried to use matrix product properties but not able to solve it.

$$ A^2+B^2=AB \quad \implies \quad (AB-BA)^2=O $$

Sangchul Lee
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Ahmed
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    An observation: these equations are invariant under similarity transformations (change-of-basis). – Jacob Manaker Jun 05 '22 at 05:01
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    It suffices to consider the case where $A$ and $B$ do not commute. Then note that Cayley–Hamilton theorem and the assumption together yield $$ \require{cancel} -BA = (A - B)^2 = \cancelto{0}{\operatorname{tr}(A-B)}(A - B) + \det(A - B)I = \det(A - B)I. $$ In order that $A$ and $B$ do not commute, the RHS must be zero, and so, we get $BA = 0$. This then implies $\det(AB - BA) = \det(AB) = \det(BA) = 0$, and so, by the Cayley–Hamilton theorem again, $$(AB - BA)^2 = \operatorname{tr}(AB - BA)(AB - BA) - \det(AB - BA)I = 0. $$ – Sangchul Lee Jun 05 '22 at 05:06
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    Why $\operatorname{tr}(A-B)=0$? – Binxu Wang 王彬旭 Jun 05 '22 at 05:12
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    $AB-BA$ is singular and traceless. Now use Cayley-Hamilton to finish. – user1551 Jun 05 '22 at 05:33
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    @BinxuWang王彬旭, You are right. I am not sure what led me to think the trace of $A - B$ is zero... – Sangchul Lee Jun 05 '22 at 06:09
  • Thanks a lot guys,Your efforts are highly appreciated. – Ahmed Jun 05 '22 at 07:06
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    Is there an example where $AB-BA$ is not already the zero matrix? It surely can be done in the complex domain, but with real matrices? – Lutz Lehmann Jun 05 '22 at 07:46

2 Answers2

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First we can use Cayley-Hamilton for a 2x2 matrix to substitute the square $$ (AB-BA)^2=(\operatorname{tr}(AB-BA))(AB-BA)-\det(AB-BA)I $$ Since $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, $\operatorname{tr}(AB-BA)=0$.

For our goal, it suffices to show that $\det(AB-BA)=0$.

For this there is a pretty clever construction proof here

Just to recap that proof, let $\omega=e^{i2\pi/3},\omega^3=1$ be 3rd unit root. Then we have this identity using $A^2+B^2=AB$ $$ \omega(AB-BA)=-(1+\omega^2)AB -\omega BA\\ =-(A^2+B^2) -\omega^2AB-\omega BA\\ =-(A+\omega B)(A+\omega^2B)\\ =-(A+\omega B)(A+\bar\omega B) $$ Thus we have a formula for the determinant $$ \det(A+\omega B)\det(A+\omega^2B)=\det(\omega(BA-AB)) $$ Since $\omega^2=\bar\omega$, $A,B$ are real, $\bar{A+\omega B}=A+\omega^2 B$. Thus LHS is a real number. $$ \det(A+\omega B)\det(A+\omega^2B)=|\det(A+\omega B)|^2 $$ While on RHS, when we deal with 2x2 matrices $$ \det(\omega(BA-AB))=\omega^2\det(BA-AB) $$ $\omega^2$ has a non-zero imaginary part, thus $\det(BA-AB)=0$. Then we prove $(AB-BA)^2=0$ by Cayley-Hamilton.

This answer is inspired by @user1551 and @Sangchul Lee :D

Suzu Hirose
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Binxu Wang 王彬旭
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If $v$ is an eigenvector of $B$ with eigenvalue $λ$, then $A^2v+λ^2v=λAv$, $$ (A-λωI)(A-λ\bar ωI)v=0,~~ ω^2-ω+1=0, $$ so that $v$ is also an eigenvector of $A$, with eigenvalue $λω$ or $λ\bar ωI$. If $B$ has 2 (strict) eigenvectors, then $A$ and $B$ are therefore simultaneously diagonalizable, thus commute, thus already $AB-BA$ is zero.

In the other case $B$ has a repeated eigenvalue which has to be real, with a real eigenvector. There is then a real basis change so that in the new basis $B=\pmatrix{λ&1\\0&λ}$, $λ$ is real with eigenvector $e_1$. The likewise changed matrix $A$ is still real and consequently has, by the above observation, the pair of conjugate eigenvalues $λω$, $λ\bar ω$, one of them with eigenvector $e_1$. However, that is impossible, as $Ae_1$ is real and at the same time equal to the strictly complex $λωe_1$.

Lutz Lehmann
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