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I was trying to solve the following problem:

Assume $A,B\in M_n\left( \mathbb{C} \right)$,satisfy $$AB^2A=AB.$$

I need to proof $$\left( AB \right) ^2=AB.$$

I tried to use some equivalent substitution of matrices, but I did not succeed. I also tried to find some counterexamples of matrices, such as 2nd order matrices, but I did not succeed either.

I don't know if this is a right problem or a wrong problem.

I hope to solve this problem. Thanks!

user1551
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fusheng
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3 Answers3

8

I have solved this problem！！！

Assume $I_n$ is a n-order unit matrix.

From $AB^2A=AB$,we can get $AB(BA-I_n)=0.$

So$$\mathrm{rank}(AB)+\mathrm{rank}(BA-I_n)\leqslant n.$$

Because $\mathrm{rank}(BA-I_n)=\mathrm{rank}(AB-I_n)$, we can get $$\mathrm{rank}(AB)+\mathrm{rank}(I_n-AB)\leqslant n.$$

Because $$\mathrm{rank}(AB)+\mathrm{rank}(I_n-AB)\geqslant \mathrm{rank}(AB+I_n-AB)=n,$$

we can get $$\mathrm{rank}(AB)+\mathrm{rank}(I_n-AB)=n.$$

Thus from the equivalent condition for idempotent matrixs, we can get $$(AB)^2=AB.$$

fusheng
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    Would this also hold in any unital associative ring? If not, can one identify the crucial property of matrices enabling this fact? – Ruy Jul 16 '22 at 19:29
  • To put it in another way': does this hold in $B(H)$, the algebra of bounded operators on Hilbert's space? – Ruy Jul 16 '22 at 20:15
  • Why is $rank(BA - I_n) = rank(AB - I_n)$? I notice also that $AB(BA-I_n) = 0$ implies $AB(BA-I_n)v = 0v = 0$ $\forall v$, so in particular the range of $BA-I_n$ is the nullspace of $AB$, hence $rank(BA-I_n) = nullity(AB)$, so by the rank-nullity theorem, we get $rank(AB) + nullity(AB) = n$ without need for double inequalities. Does this reasoning go backwards to say $rank(I_n - AB) = nullity(AB)$ implies the range of $(I_n-AB)$ is the nullspace of $AB$? So that we may conclude $AB(I_n-AB) = 0$ as in your proof? – I Zuka I Jul 17 '22 at 11:01
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Fusheng's own answer is elegant. Here I am just trying to find an alternative solution.

Using Fitting's decomposition or Jordan form, we may write $\mathbb C^n=V\oplus W$ where $V$ and $W$ are two invariant subspaces of $BA$ such that $BA$ is nonsingular on $V$ and nilpotent on $W$. Let $(BA)^mW=0$. Then $$ ABW=ABBAW=(ABBA)BAW=AB(BA)^2W=\cdots=AB(BA)^mW=0. $$ Therefore, by a change of basis, we may assume that $$ AB=\pmatrix{X&0\\ Z&0}\quad\text{and}\quad BA=\pmatrix{M\\ &N} $$ where $M$ is nonsingular, $N$ is nilpotent and $X$ has the same size as $M$. Since $AB$ and $BA$ have the same characteristic polynomial, $X$ and $M$ have the same spectrum. Hence $X$ is nonsingular and the condition $ABBA=AB$ implies that $XM=X$ and in turn $M=I$. Thus $X$ is unipotent (because it has the same spectrum as $M$). However, as $AB-I$ and $BA-I$ have the same rank, we must have $X=I$. Therefore $AB=\pmatrix{I&0\\ Z&0}$ is idempotent.

user1551
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So we know that $ABBA=AB$. Supposing that the inverse elements $A^{-1}$ and $B^{-1}$ exist, we can multiplity this from the left with $A^{-1}$ to get $BBA=B$ and then multiply with $B^{-1}$ from the left to get $BA=1$. Insert this into the product $AB^2A=ABBA$ to give $AB(BA)=AB1=AB$ which was to be shown.

R. J. Mathar
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    $A$ and $B$ are not specified as invertible matrices, so we can't assume their inverses exist. –  Jul 16 '22 at 14:11