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Let $T$ an operator from infinite dimensional vector space to infinite dimensional vector space .

Is every eigenvalue spectrum point or is every spectrum point eigenvalue or there is relation between them ?

I would like to recieve the answer with example please ?

Let $T$ from $l^2$ to $l^2$ such that $x=(x_1,x_2,x_3,.....)$ then $Tx=(x_1/1,x_2/2,x_3/3,....)$ how can I find the eigenvalues point spectrum for $T$

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    What's the definition of spectrum? What's the definition of an eigenvalue? – 2017-02-07
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    Some examples (especially an operator with spectrum, but no eigenvalues) can be found here: http://www-users.math.umn.edu/~garrett/m/fun/notes_2012-13/06b_examples_spectra.pdf – 2017-02-07
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    I think you need more structure on the spaces (e.g. norm, inner product) to determine a relationship between point spectrum and collection of eigenvalues. – 2017-02-07
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    The operator in your example is compact and self-adjoint, which gives a lot of nice structure your general question does not specify. – 2017-02-07

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Let's extend on your example:

Let $n \in \mathbb N$ and $e_n$ be the $\ell^2$-series which is equal to $1$ at the $n$th place and equal to $0$ everywhere else, e.g. $e_1=(1,0,\dots), e_2=(0,1,0,\dots)$ and so on.

Then $Te_n = (0,0,\dots,\frac{1}{n}\cdot1,0,\dots)= \frac{1}{n}e_n,$ which means that $\frac{1}{n}$ is an eigenvalue with the eigenvector $e_n$, i.e. it is part of the point spectrum of $T$.

Thus, the spectrum of $T$ contains at least $\{1,\frac{1}{2},\frac{1}{3},\dots\}=\{\frac{1}{n}, n \in \mathbb N\}$. Since the spectrum of $T$ is closed (why?), also accumulation points of the spectrum are part of the spectrum. In this case, $\lim_{n \rightarrow \infty} \frac{1}{n}=0$.

So, we have that $0$ is a point of the spectrum of $T$. But is it an eigenvalue, i.e. a member of the point spectrum? If it were, there would be some nonzero $x =(x_1,x_2,\dots) \in \ell^2$ such that $Te = 0\cdot e = 0$, i.e.

$$Tx=(x_1,\frac{x_2}{2}, \frac{x_3}{3}, \dots) = (0,0,0,\dots),$$

which means that $x_1=x_2=x_3=\dots=0,$ which shows that there cannot exist an eigenvector, thus $0$ is not an eigenvalue.

To restate: $\sigma(T)\supset \{\frac{1}{n}:n \in \mathbb N\}\cup \{0\}$ (in fact, it's equal, but that beyond the scope of this answer) and $\{\frac{1}{n}:n \in \mathbb N\}$ are eigenvalues (i.e. members of the point spectrum) and $0$ is a spectral point, but not a member of the point spectrum.

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    There is a small error in your last paragraph: what you showed was actually the reverse inclusion, that is, $\{\frac{1}{n}:n \in \mathbb N\}\cup \{0\}\subset\sigma(T)$. – 2017-02-07