matrices such that $A^2B+BA^2=2ABA$

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

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}

• Would you mind giving me an elementary proof for Jacobson’s Lemma, please! Mar 27, 2021 at 21:02

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 $$\Bbb C^{3\times3}$$.

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!

• nice proof @Robert Mar 27, 2021 at 14:00
• Isn’t the final step wrong? I think the given shows that $AC-CA$ is zero instead (that is, that $A$ and $C$ commute). Mar 27, 2021 at 14:14
• @mjqxxxx is right. Also, I'm not sure about the reason he gave for believing that they commute.
– Anon
Mar 27, 2021 at 14:38
• 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. Mar 27, 2021 at 15:23