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Let $X$ be a vector space over $\mathbb{C}$. The set $\mathcal{L}(X)$ of all linear mappings $A: X \to X$ is a vector space over $\mathbb{C}$. Prove, or disprove with a counterexample, the following assertion:

\begin{align*} \rho(A) = \textrm{max}\{|\lambda| \in \mathbb{R} : \lambda \textrm{ is an eigenvalue of $A$}\} \end{align*} defines a norm on $\mathcal{L}(X)$.

It seems to me that this is not a norm. For a counterexample, consider any nonzero nilpotent operator $N$. Since $N$ is nilpotent, its only eigenvalues are 0, hence $\rho(N) = 0$. However, for $\rho$ to be a norm, $\rho(X) = 0$ if and only if $X = 0$, so $\rho$ is not a norm. Is there something I'm missing here?

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    $\begingroup$ You make a good point. $\endgroup$
    – hardmath
    Jan 2 at 4:49
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    $\begingroup$ Of course, the question is if such a non-zero nilpotent operator exists. If you can give one, that would be the counterexample. $\endgroup$
    – Ingix
    Jan 2 at 9:53
  • $\begingroup$ Good point. Would something like this work: Let $T \in \mathcal{L}(X)$ be a nonzero operator on $X$. Since $X$ is a vector space over $\mathbb{C}$, there exists a decomposition, $T = D + N$, where $D$ is a diagonalizable matrix and $N$ is a nilpotent matrix, $N$ is a suitable counterexample. $\endgroup$ Jan 3 at 2:30
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    $\begingroup$ Your $N$ could be zero in that example and it always will be if $\operatorname{dim} X = 1$ (in fact in this case the map is a norm). For $\operatorname{dim} X > 1$ take e.g. two linearly independent vectors $v_1, v_2$ and define $N$ by $v_1\mapsto v_2, v_2\mapsto 0$ and extend $N$ by zero on a complement of $\operatorname{span} v_1,v_2$. $\endgroup$
    – leoli1
    Jan 3 at 18:30
  • $\begingroup$ I see how $v_1 \mapsto v_2, v_2 \mapsto 0$ works, but I don't understand what you mean by "extend $N$ by zero on a complement of span $v_1,v_2$. $\endgroup$ Jan 3 at 19:44

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