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How to prove that $AB$ is a projection if $(AB)(BA)=AB$?

The above question was asked by me, and I found the solution by myself.

From the above problem, we can know that for $A,B\in M_n\left(\mathbb{C}\right)$, if $AB^2A$= $AB$,

we can get $$(AB)^2=AB.$$

Thus I have the following problem:

does this also hold in $\mathscr{B}(\mathcal{H})$, the algebra of bounded linear operators on Hilbert's space?

That is for any $A,B \in \mathscr{B}(\mathcal{H})$, if $AB^2A=AB$,

can we get $$(AB)^2=AB?$$

Thanks.

fusheng
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1 Answers1

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Oh,I also find the solution about the problem.

Let $$\mathcal{H}=l^2.$$

Let $A(x_1,x_2,\cdots)=(0,2x_1,x_2,\cdots)$ ,thus we can get $||A||=2.$

Let $B(x_1,x_2,\cdots)=(x_2,\cdots)$ , thus we can get $||B||=1.$

We also get $$ABBA (x_1,x_2\cdots)=(0,2x_2,x_3,\cdots)$$

and $$AB (x_1,x_2\cdots)=(0,2x_2,x_3,\cdots).$$

So for the above $A$ and $B$ , we can get $$ABBA=AB.$$

But $$(AB)^2(x_1,x_2\cdots)=(0,4x_2,x_3\cdots).$$

So for the above $A$ and $B$, we can get $$(AB)^2\ne AB.$$

fusheng
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  • Very clever! Congratulations! – Ruy Jul 21 '22 at 03:07
  • The next case to be considered is a finite von Neumann algebra. Are you familiar with this concept? – Ruy Jul 21 '22 at 03:17
  • @ Ruy Sorry, I am not familiar with this concept. – fusheng Jul 21 '22 at 03:32
  • Ok, so here is a simplified definition. Consider a $$-subalgebra $M\subseteq B(H)$ ("$$" means closed under taking adjoints), which admits a continuous linear functional $\tau:M\to \mathbb C$, such that: (1) $\tau(ab)=\tau(ba)$ (this is the trace property), (2) $\tau(a^a)\geq0$ (positivity property), and (3) $\tau(a^a)=0\Rightarrow a=0$ (faithfulness property). Then try to solve your problem within $M$. To see the relevance of these hypotheses, try to prove that the shift, i.e. your operator $B$, can't be inside such an algebra. The hint is that $BB^=I$, but $B^B\neq I$. – Ruy Jul 21 '22 at 04:38
  • @ Ruy Thanks! I know what you are talking about. For the finite von Neumann algebra, the above example may no longer fit. So we may need to find another counterexample or give a proof if this proposition is true for the finite von Neumann algebta. But I am not familiar with the finite von Neumann algebra. – fusheng Jul 21 '22 at 09:33