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.
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 $$(AB)^2\ne AB.$$