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Let be $A$, $B$ two matrices $3 \times 3$ with complex entries. Prove that if $$A^2B+BA^2=2ABA$$ THEN $$B^2A+AB^2=2BAB$$ I tried it and do not know how to continue. If $A$ is invertible then $$AB^2+A^{-1}BA^2B=2BAB$$ so I have to prove that $$B^2A=A^{-1}BA^2B$$ How to continue and also have to discuss the case when $A$ is not invertible

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The statement in question is equivalent to $[A,[A,B]]=0\rightarrow[[A,B],B]=0$ and it is false. Here is a counterexample taken from the last section of Irving Kaplansky, Jacobson's Lemma Revisited, Journal of Algebra, 62, 473-476 (1980): \begin{aligned} A&=\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1},\quad B=\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0},\\ [A,B]&=\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1}\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}-\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1}\\ &=\pmatrix{0&0&1\\ 0&0&1\\ 1&0&0}-\pmatrix{0&0&0\\ 0&0&1\\ 1&1&0} =\pmatrix{0&0&1\\ 0&0&0\\ 0&-1&0},\\ [A,[A,B]]&=A[A,B]-[A,B]A=[A,B]-[A,B]=0,\\ [[A,B],B]&=\pmatrix{0&0&1\\ 0&0&0\\ 0&-1&0}\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}-\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}\pmatrix{0&0&1\\ 0&0&0\\ 0&-1&0}\\ &=\pmatrix{1&0&0\\ 0&0&0\\ 0&0&-1}-\pmatrix{0&0&0\\ 0&-1&0\\ 0&0&1} =\pmatrix{1&0&0\\ 0&1&0\\ 0&0&-2}\ne0. \end{aligned}

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  • $\begingroup$ Would you mind giving me an elementary proof for Jacobson’s Lemma, please! $\endgroup$
    – shangq_tou
    Mar 27, 2021 at 21:02
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We'll work with commutators, $[X,\,Y]:=XY-YX$. Given $[A,\,[A,\,B]]=O$, we want to prove $[[A,\,B],\,B]=O$. Since $A,\,[A,\,B]$ are simultaneously diagonalisable, we need only show any base diagonalising both - we'll work hereafter in such a base - also diagonalises $B$. The easiest case is one in which $A$ has no repeated eigenvalues: if $i\ne j$,$$0=[A,\,B]_{ij}=(\underbrace{A_{ii}-A_{jj})}_{\ne0}B_{ij}\implies B_{ij}=0.$$The general case follows by such "nondegenerate" $A$ being dense in $\Bbb C^{3\times3}$.

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If $A$ and $B$ commute, the result is immediate.

Note that $(A+B)^3 = A^3 + (A^2B + BA^2) + ABA + (B^2A + AB^2) + BAB + B^3$. If the result holds, that turns into $(A+B)^3 = A^3 + A^2B + AB^2 + B^3$. This suggests that $A$ and $B$ might have to commute for the given to hold.

If $A$ and $B$ do not commute, let $AB-BA = C$. Rewrite that as $BA = AB - C$ and as $AB = BA + C$. Plugging in, we get

$$B^2A = B(AB-C) = BAB-BC$$and $$AB^2 = (C+BA)B = CB + BAB$$ so $$B^2A + A^2B = 2BAB - BC + CB$$ without even using the given. Turning to that, we get $$A^2B + BA^2 = 2ABA + AC-CA$$ by the same reasoning. This implies $AC-CA = 0$. So now our goal is to get from $AC=CA to BC=CB.$. But according to the other answer by @user1551, the claim isn't true in the first place!

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  • $\begingroup$ nice proof @Robert $\endgroup$ Mar 27, 2021 at 14:00
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    $\begingroup$ Isn’t the final step wrong? I think the given shows that $AC-CA$ is zero instead (that is, that $A$ and $C$ commute). $\endgroup$
    – mjqxxxx
    Mar 27, 2021 at 14:14
  • $\begingroup$ @mjqxxxx is right. Also, I'm not sure about the reason he gave for believing that they commute. $\endgroup$
    – Kaind
    Mar 27, 2021 at 14:38
  • $\begingroup$ Thanks for the correction! In my defense, I was tilting at a windmill anyway...but I should have caught that. Note to self: Write out every step, Rob. $\endgroup$ Mar 27, 2021 at 15:23

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