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