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

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.

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.
• Yes, so far I've only found the version on a finite-dimensional complex inner product space. Maybe this link will be of some use. Jun 11, 2020 at 7:44
• Thanks. Will study the material on the link. Jun 11, 2020 at 23:28

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. $$A^2=A$$.

• Looks good! Thanks :-) Jun 12, 2020 at 0:46

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 $$A^2v=Av$$.

• That's an elegant method indeed! However, it seems to me that this method won't work in case the underlying field is real. That's because, the Spectral Theorem for normal operators (in finite dimensions) holds if and only if the underlying field is complex, not real. Accordingly, an arbitrary normal operator on a finite-dimensional real inner product space is not guaranteed to be diagonalizable. Thus, we can't really conclude "$A^2v = Av$", in the finite-dimensional subspace suggested by you. Am I missing something here? Jun 11, 2020 at 23:37
• No, a normal operator that satisfies $A^2(A-I)=0$ can have only 0 and/or 1 as eigenvalues, so it is diagonalisable over the reals as well. Jun 12, 2020 at 7:20
• Yes, that's right! It was a subtle point that I missed ... Jun 13, 2020 at 1:05