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Let $A$ and $B$ be two $3\times 3$ matrices with complex entries such that $A^2=AB+BA$. Prove that $\det(AB-BA)=0$.

(Is the above result true for matrices with real entries?)

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    $\begingroup$ What reason do you have to believe that this holds? Do you have an nontrivial example? $\endgroup$
    – flawr
    Jan 26 '15 at 15:08
  • $\begingroup$ As a remark, counterexamples exist when the sizes of $A$ and $B$ are even. E.g. when $A=\pmatrix{1&-1\\ 0&-1}$ and $B=\pmatrix{1&0\\ 1&-1}$, we have $A^2=AB+BA=I$ but the determinant of $AB-BA=\pmatrix{-1&2\\ -2&1}$ is $3\ne0$. $\endgroup$
    – user1551
    Jan 26 '15 at 16:39
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Let $A,B\in F^{n,n}$, $F$ any field of characteristic $\ne 2$, $n$ odd. Assume they satisfy $A^2 = AB+BA$.

From $A^2=AB+BA$ it follows $$ AB-BA= A^2-2BA= (A-2B)A. $$ Then it holds $$ \det(AB-BA) = \det((A-2B)A) = \det(A(A-2B)). $$ Now we have $$ A(A-2B) = A^2-2AB = BA-AB. $$ This proves $$ \det(AB-BA) = \det(BA-AB) = \det( -(AB-BA)) = (-1)^n \det(AB-BA), $$ and since $n$ is odd (and $1+1\ne0$), $\det(AB-BA)=0$ follows.


If the field has characteristic 2 then the result is true as well: In this case $AB-BA=AB+BA=A^2$. Suppose $A$ is invertible. Then the assumption $A^2=AB+BA$ implies $I_3=BA^{-1}+A^{-1}B$. The trace of the matrix on the left is $1$, the trace of the matrix on the right is $tr(BA^{-1}+A^{-1}B)=tr(BA^{-1})+tr(BA^{-1})=0$, contradiction. So $A$ cannot be invertible, and $\det(AB-BA)=\det(A^2)=0$.

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  • $\begingroup$ In a field of characteristic two, the last step fails. Indeed, from $D = -D$, we get $2 \, D = 0$ and this is not enough to conclude $D = 0$. $\endgroup$
    – gerw
    Nov 2 '20 at 13:44
  • $\begingroup$ @gerw thanks for the hint. In fact, the claim is also true in this case (with a different and simplier proof) $\endgroup$
    – daw
    Nov 2 '20 at 15:01
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This was easier than I originally thought. Note that $$ AB - BA = AB + BA - 2BA = A^2 - 2BA = (A-2B)A\\ -(AB - BA) = AB + BA - 2AB= A^2 - 2AB = A(A - 2B) $$ and that we have both $\det(AB - BA) = \det(-(AB - BA))$ and $\det(AB - BA) = -\det(-(AB - BA))$.


My original approach: note that $AB - BA$ has trace zero. Then, we note that $$ AB - BA = AB + BA - 2BA = A^2 - 2BA = (A-2B)A\\ -(AB - BA) = AB + BA - 2AB= A^2 - 2AB = A(A - 2B) $$ From there, we note by Sylvester's determinant theorem that $$ \det(AB - BA - \lambda I)= \\ \det((A - 2B)A - \lambda I) =\\ \det(A(A - 2B) - \lambda I) =\\ \det(-(AB - BA) - \lambda I) =\\ (-1)^3 \det(AB - BA + \lambda I) $$ So, if $\lambda$ is an eigenvalue of $AB - BA$, then so is $-\lambda$. (Alternatively, directly note that $(A - 2B)A$ and $A(A - 2B)$ must have the same eigenvalues).

Thus, we know that $AB - BA$ has $3$ eigenvalues whose sum is zero and such that $\lambda$ is an eigenvalue iff $-\lambda$ is an eigenvalue. We conclude that $AB - BA$ must have $0$ among its eigenvalues. That is, $\det(AB - BA - 0I) = \det(AB - BA) = 0$.

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