Infinite-dimensional inner product spaces: if $A^3 = A^2$ for a normal operator $A$, then $A^2 = A$

Question

Paul R. Halmos, Finite-Dimensional Vector Spaces, sec. 80, p.162, exercise 5(c):

If $A$ is normal and $A^3 = A^2$, then $A$ is idempotent.

The finite-dimensional case has been asked before, but our $A$ here is a linear operator on a possibly infinite-dimensional inner product space. The underlying field may be real or complex.

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My attempt so far: I see that the finite-dimensional version of this problem (over a complex field) is easy to address using the Spectral Theorem for normal operators (on complex inner product spaces). To prove the assertion in infinite dimensions, my attempts so far have been around showing that the "distance" between the vectors $A^2x$ and $Ax$ (if $x$ is an arbitrary vector) is zero, i.e., $\Vert A^2x-Ax\Vert = 0$. Haven't been successful. Would appreciate help. Thanks.

Answer 1

In general, if $T^\ast T=0$ for some linear operator $T$, then $\langle Tx,Tx\rangle=\langle T^\ast Tx,x\rangle=0$ for all $x$. Hence $Tx=0$ for all $x$, i.e. $T=0$.

Now, since $A$ is normal, $B=A^2-A$ is normal. As $A^2=A^3$, we have $A^2=A^3=A(A^2)=A(A^3)=A^4$. Therefore $B^2=0$. It follows that $(B^\ast B)^\ast(B^\ast B)=(B^\ast)^2B^2=0$. Hence $B^\ast B=0$. In turn, $B=0$, i.e. $A^2=A$.

Answer 2

This is going to sound familiar :)

Let $v$ be any vector, form the subspace generated from $v,Av,A^2v$; it is $A$-invariant since $A^3=A^2$. So, from the finite-dimensional case, it follows that $A^2v=Av$.