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I think that the spectrum of a nilpotent operator is {0} but I don't know how to prove it and that's just an intuition. I think that for b), it's wrong because the only diagonalisable nilpotent matrix is the zero matrix but again I think I need a counterexample here. But I might be making a mistake here, is a nilpotent matrix a nilpotent operator ? enter image description here

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    $\begingroup$ According to this answer, you are right about the spectrum of a nilpotent operator containing only zero. $\endgroup$ Mar 29, 2016 at 19:38

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is a nilpotent matrix a nilpotent operator ?

A nilpotent operator $f: V \rightarrow V$ can be represented as a matrix through one (or two, if you feel like being tricky) base(s) of $V$. Generally, an operator can be represented as a matrix, as long as a base is provided. If a matrix $M$ represents a nilpotent operator $f$ so that $f^n = 0$, then it satisfies $M^n = 0$. Think of a matrix as one way of representing an operator. Crappy metaphor incoming: "hola" and "hello" both represent the same thing, greeting, in two different languages. Here, the concept of "greeting" plays the role of the operator, those two words are two representations (playing the role of matrices) of this operator in two different bases, respectively, the Spanish and English languages. Not sure if that's any help.

Proofs of both questions

(a) Your intuiton is right. Let's prove $Sp(f) = \{0\}$ with double inclusion.

First: let's prove, through reductio ad absurdum, that $Sp(f) \subset \{ 0 \}$. Suppose $\lambda \neq 0$ is in $f$'s spectrum. Let $u \in V$ be an eigenvector of $\lambda$. $f^n(u) = \lambda^n u \neq 0$, which contradicts $f^n \neq 0$. Hence, $Sp(f) \subset \{0 \}$.

Second: let's (directly) prove that $0$ is necessarily in $Sp(f)$. If $f=0$, that goes without saying. Else, define $m$ as the smallest integer so that $f^m = 0$. $f^{m-1} \neq 0$, therefore, there exists some $u \in V$ so that $f^{m-1}(u) \neq 0$. Any of these non-zero $f^{m-1}(u)$ is an eigenvector for the $0$ eigenvalue. Hence, $0$ is in $Sp(f)$.

(b) Given that $Sp(f) = \{0\}$ seems true, having (b) also true would mean any nilpotent operator is similar to $0$, which sounds weird. Let's find a counter-example: a nilpotent operator that is not diagonalizable. A quick Wikipedia search reveals this simple example:

$ C := \left[ \begin{array}{cc} 0 & 1 \\ 0&0\end{array} \right] $

$C$ is nilpotent ($C^2=0$), yet it is not diagonalizable (not over $\mathbb{R}$ nor over $\mathbb{C}$).

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  • $\begingroup$ And what happens if there are no eigenvectors corresponding to the eigenvalue $\lambda$ (imposssible in finite-dimensioned case, yet possible in infinite-dimensioned case)? $\endgroup$ Mar 29, 2016 at 20:40
  • $\begingroup$ Thank you very much, that's what I thought ! However, one of my question was is a matrix like C an operator ? It seems that yes. $\endgroup$
    – TedMosby
    Mar 29, 2016 at 21:17
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    $\begingroup$ @TZakrevskiy good call. Hadn't thought about it... but my gut feeling is that the person who wrote the question would accept such an answer. I may be wrong. $\endgroup$
    – Charles
    Mar 30, 2016 at 7:04
  • $\begingroup$ @ThéodoreRozencwajg C is a matrix. It represents an operator in a given base of $V$. By abuse of terminology, it could be said that "C is an operator", but technically it is not. However, a lot of properties of matrix or of operators (including nilpotence) are defined so that if it is true for a given operator, it is also true for any of its representations as a matrix. $\endgroup$
    – Charles
    Mar 30, 2016 at 7:09

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