# Invertibility of linear operators on infinite dimensional vector space

Let $E$ be a finite dimensional vector space, and $T$ a bounded linear operator from $E\rightarrow E$. In this case:

$$T \text{ invertible} \leftrightarrow \text{null } T = \{0\}$$

In an infinite dimensional vector space, this is not the case, but I am struggling to see how that is true. Can you help me find an example of an invertible operator that has a non-trivial nullspace? Or if that does not exist, then an operator with a trivial null space that is not invertible?

For some context, I am trying to better understand the difference between spectrum of an operator and its eigenvalues--and this seems to be the central issue.

An invertible operator can't have a non zero null space since the image of the null space is zero therefore a non zero null space implies that the operator is not injective.

Consider the shift operator defined by $T(e_n)=e_{n+1}$ in $Vect(e_0,..,e_n,...)$ its image does not contains $e_0$, so it is not surjective.

• So, is it safe to say that in infinite dimensional space, a trivial null space gives you injectivity, but not necessarily surjectivity. So all examples of non invertible operators with trivial null space must not be surjective? Commented Apr 30, 2017 at 11:28
• yes, all examples of non invertible operators with trivial null space are not surjective. Commented Apr 30, 2017 at 13:09
• Thank you, this is very helpful. Perhaps it should be a separate question, but how does the situation change if E is infinite dimensional, but T is a compact operator? Notably, the example you gave is not compact, and since compact operators are the limit of finite rank operators, it seems plausible that this is a relevant condition. Commented Apr 30, 2017 at 14:51