Let $x$ be an operator in $B(H)$. By definition $\sigma(x)=\{\lambda \in \Bbb C ~; \lambda - x \neq inv \}$. Also to find eigenvalue of an operator we should find $\lambda$ such that $x\xi = \lambda \xi$ for some $\xi\in H$. In this case we show that $\lambda - x \neq inv$, so every eigenvalue belongs to spectrum. I know operators that their spectrum do not contain any eigenvalue as bilateral shifts. While I think the process of finding spectrum is the same as eigenvalue. I think to show $x-\lambda$ is not invertable, we should find $\xi \in H$ such that $x\xi = \lambda\xi$ and it shows that $\sigma (x)$ is a subset of eigenvalues set of $x$, which is not true in general. Thus I think we should find spectrum in another way. Please help me to understand it. Thanks in advance

  • $\begingroup$ @user7530 : inv means invertable. As I mentioned spectrum of bilateral shiftes contain no eigenvalues. Also right shift operators. $\endgroup$ – niki Dec 28 '14 at 6:56
  • $\begingroup$ Yes, you're right. The spectrum will contain all eigenvalues, and also possible some additional elements that aren't eigenvalues. $\endgroup$ – user7530 Dec 28 '14 at 7:03
  • $\begingroup$ @user7530 : I want to know how we can find additional elements $\endgroup$ – niki Dec 28 '14 at 7:04

There are basically two ways a bounded operator $T$ can fail to be invertible:

  1. Its null space is nontrivial (that leads to eigenvalues/eigenvectors), or
  2. Its range is not the whole space.

Condition 2 can also be divided into two cases:

2a. The null space of $T^*$ is nontrivial, i.e. there is a nonzero vector orthogonal to the range of $T$

2b. The range of $T$ is not closed.

If the range is not closed but the null space is trivial, $$\inf_{0 \ne v \in H} \dfrac{\|Tv\|}{\|v\|} = 0$$

  • $\begingroup$ you mean that $\inf_{\|v\|=1}\|Tv\| =0$, but I do not know why you claim it. $\endgroup$ – niki Dec 28 '14 at 7:38
  • 1
    $\begingroup$ If $\text{ker}\; T = \{0\}$, $\text{ran}\; T$ is closed iff the map $Tv \to v$ is bounded (use the Open Mapping Theorem). $\endgroup$ – Robert Israel Dec 28 '14 at 9:52

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