# Max Eigenvalue Norm

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?

• You make a good point. Jan 2 at 4:49
• Of course, the question is if such a non-zero nilpotent operator exists. If you can give one, that would be the counterexample. Jan 2 at 9:53
• 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. Jan 3 at 2:30
• 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 $\operatorname{span} v_1,v_2$. Jan 3 at 18:30
• 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 $v_1,v_2$. Jan 3 at 19:44