Exercise of compact self-adjoint operator

Question

'Let $H$ be a Hilbert space. Find all compact self-adjoint operators $T:H \rightarrow H$ such that $T^{k}=0$ with $k>0, k \in N$.' $ \ $

I have this idea. Consider $\lambda_n$ eigenvalue of T and $e_n$ its corresponding eigenvector. Then $Te_n=\lambda_n e_n $. So: $T^{k}e_n=T^{k-1}(Te_n)=\lambda_nT^{k-1}e_n=...=\lambda_n^{k}e_n =0$. And we have this for all eigenvalues and eigenvectors. So $T$ should be $T=0$?

Answer 1

If $T$ is self-adjoint and $T^2x=0$, then $Tx=0$ because $$ \|Tx\|^2=\langle T^2 x,x\rangle = 0. $$ Therefore, if $T^k=0$ for some $k > 1$, then $T=0$. This doesn't rely on $T$ being compact, but it does rely on $T$ being self-adjoint.

Answer 2

Your argument is correct, but showing that the spectrum is $\{0\}$ does not in general imply that $T=0$; it does, though, when $T$ is selfadjoint but you need to include that argument.

The two usual ways to go would be

• use the Spectral Theorem, that expresses $T$ (since it is compact and selfadjoint) in terms of its eigenvalues.

• Use the formula for the spectral radius. You have that $$\tag1 \operatorname{spr}(T)=\lim_n\|T^n\|^{1/n}. $$ In your setup, this shows that $\sigma(T)=\{0\}$. But to conclude that $T=0$ you need to mix this with the fact that $T=T^*$. You have $\|T^2\|=\|T^*T\|=\|T\|^2$. Using induction you get $\|T^{2n}\|=\|T^{2n}\|$. Now using $(1)$ you have $$ 0=\lim_n\|T^n\|^{1/n}=\lim_n\|T^{2n}\|^{1/2n}=\|T\|. $$ So $\|T\|=0$ and then $T=0$. This last argument doesn't use that $T$ is compact, so it shows that any selfadjoint operator with spectrum $\{0\}$ is equal to zero.