Invariant subspaces for this linear extension of operators 
Let $(e_k)$ be a total orthonormal sequence in a separable Hilbert space
  $H$ and let $ T: H\to H$ be defined at $e_k$ by
$T(e_k)=e_{k+1}$ , $(k=1,2,\cdots)$
and then linearly and continuously extended to $H$. Find invariant
  sub-spaces.
Show that $T$ has no eigenvalues.

I know that the answer is $X_n=\operatorname{span}\{e_n,e_{n+1},\dots\}$are invariant subspaces which I got from E.Kreyszig's Introductory Functional Analysis book. But I can't understand that how to prove the result. Also I'm unable to find any clue to prove the second result. 
Can someone help me?
Thanks in advance...
 A: Assume $x$ is an eigenvector corresponding to an eigenvalue $\lambda$. We have $x = \sum_{i=1}^\infty x_ie_i$, where $x_i = \langle x,e_i\rangle$. Let us start with $\lambda = 0$. Then $Tx = 0$. Hence,
$$
0 = T\sum_ix_ie_i = \sum_i x_iTe_i = \sum_i x_ie_{i+1}.
$$
But this implies $x_i = 0$ for all $i$ and so $x = 0$. Can you do a similar "trick" when $\lambda\neq 0$?
A: It's easy to see that you spaces are invariant by $T$, in fact If we put $S_k=\overline{Vect\{(e_i)_{i\geq k}}\}$ then $S_k$ is a closed subspaces of H and for all $x\in S_k$ it exist a sequence $(a_i)_{i\geq k}$ of complexe numbers in $l^2$ such that :
$$
x=\sum_{i\geq k} a_i e_i \qquad (x=(a_1,a_2,a_3,\dots))
$$
Then 
$$
Tx=T\left( \sum_{i\geq k} a_i e_i\right)=\sum_{i\geq k} a_i Te_i=\sum_{i\geq k} a_i e_{i+1} \qquad (x=(0,a_1,a_2,a_3,\dots))
$$
This Operator is called unilateral Shift. the set of all invariant subspace (Latices of sub-spaces) is known using the theory of complex functions, and it's called Beurling–Lax theorem 
