3
$\begingroup$

'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 $T=0$?

$\endgroup$
4
  • 1
    $\begingroup$ You have proved that that eigen values are $0$ but that doesn't imply that $T=0$. $\endgroup$ May 21 '20 at 7:52
  • $\begingroup$ 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 $T^k=0$? $\endgroup$
    – siteb
    May 21 '20 at 8:01
  • $\begingroup$ There isn't a $T$ other then $0$ with $T^{k}=0$. So $T=0$. $\endgroup$ May 21 '20 at 8:03
  • $\begingroup$ 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. $\endgroup$ May 21 '20 at 8:14
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ How does this show that if $T^3=0$ then $T=0$? $\endgroup$ May 21 '20 at 20:23
  • $\begingroup$ @MartinArgerami : If $T^3x=0$, then $T^2(Tx)=0$, which implies $T(Tx)=0$, which implies $Tx=0$. $\endgroup$ May 22 '20 at 4:44
  • 1
    $\begingroup$ Nice. $ \ \ \ \ $ $\endgroup$ May 22 '20 at 4:47
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.