5
$\begingroup$

I have to prove that if $\lambda$ is an eigenvalue of T then $\bar\lambda$ is eigenvalue of $T^*$ (adjoint)

I know that $<Tv,u> = <\lambda v,u> = \bar\lambda<v,u>=<v,\bar\lambda u>$

and that $<Tv,u>=<v,T^*u>$

but does this imply that there is a $u$ such that $T^*u=\bar\lambda u ?$

I believe something is wrong here. Any help?

$\endgroup$
3
  • $\begingroup$ No, your arguments does not show the existence of such $u$. $\endgroup$
    – gerw
    Dec 19, 2013 at 20:40
  • $\begingroup$ Really? I think that such a $u$ exists by using the fact that the operator $T^*- \bar{\lambda}\cdot {\rm Id}$ is not surjective, and hence it has a nontrivial kernel. $\endgroup$
    – emeu
    Dec 19, 2013 at 20:46
  • $\begingroup$ Is your vector space finite dimensional? $\endgroup$
    – egreg
    Dec 19, 2013 at 20:48

1 Answer 1

Reset to default
2
$\begingroup$

If $Tv=\lambda v$, with $v\ne 0$, then, for all $u\in V$, $$\langle T^*u-\bar\lambda u,v\rangle=\langle T^*u,v\rangle-\langle\bar\lambda u,v\rangle=\langle u,Tv\rangle-\lambda\langle u,v\rangle=0 $$ This means that the image of $T^*-\bar\lambda I$ is contained in the orthogonal complement of $v$, so $T^*-\bar\lambda I$ is not surjective. What can you say, now?

Of course, the assumption is that the space $T$ operates on is finite dimensional, because the assertion is false for infinite dimensional spaces.

$\endgroup$
3
  • $\begingroup$ I don't understand: why doesn't it work in infinite dimension? We still have the fact that non-surjectivity implies non-injectivity, haven't we? $\endgroup$
    – emeu
    Dec 19, 2013 at 20:59
  • $\begingroup$ @emeu No; an operator can be injective and not surjective in infinite dimensional Hilbert spaces. Did you look at the counterexample? $\endgroup$
    – egreg
    Dec 19, 2013 at 21:00
  • $\begingroup$ Ok I just saw the counter-example, thanks! $\endgroup$
    – emeu
    Dec 19, 2013 at 21:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .