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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – The_Anomaly Apr 30 '17 at 11:28
  • $\begingroup$ yes, all examples of non invertible operators with trivial null space are not surjective. $\endgroup$ – Tsemo Aristide Apr 30 '17 at 13:09
  • $\begingroup$ 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. $\endgroup$ – The_Anomaly Apr 30 '17 at 14:51

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