# 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 $$k>0, k \in N$$.' $$\$$

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 $$T=0$$?

• You have proved that that eigen values are $0$ but that doesn't imply that $T=0$. May 21 '20 at 7:52
• Thanks! I have know this idea: There is a result that the set of non-zero eigenvalues of $T$ (self-adjoint and compact) is not empty. So that means that there isn't a $T$ with $T^k=0$? May 21 '20 at 8:01
• There isn't a $T$ other then $0$ with $T^{k}=0$. So $T=0$. May 21 '20 at 8:03
• The missing ingredient for your idea is how to relate knowledge about eigenvalues of $T$ back to the entirety of $T$ itself. Some basic results about compact self-adjoint operators should suffice. May 21 '20 at 8:14

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.

• How does this show that if $T^3=0$ then $T=0$? May 21 '20 at 20:23
• @MartinArgerami : If $T^3x=0$, then $T^2(Tx)=0$, which implies $T(Tx)=0$, which implies $Tx=0$. May 22 '20 at 4:44
• Nice. $\ \ \ \$ May 22 '20 at 4:47

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.