Counterexample: linear operator on infinite dimensional space with no eigenvector

[Bachelier]

Can you guys provide a counter example of why this statement is False.

If T: V-->V with V a VS over C then T has an eigenvector?

This is not always true as if V is infinite dim., it'll have a Spectrum.

Any counter examples?

Thanks

 

[Landau]

The standard example is the bilateral shift on $\ell(\mathbb{Z})$:

$$T:\ell(\mathbb{Z})\to \ell(\mathbb{Z})$$
$$(a_i)_{i\in\mathbb{Z}}\mapsto (a_{i+1})_{i\in\mathbb{Z}}.$$