Suppose we have a Jordan block with dimension greater than $2$. For example, let us denote \begin{align*} J = \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix}. \end{align*} I am wondering whether there exists some vector norm $\| \cdot \|_v$ (the subscript denotes some vector norm) such that the induced operator norm of $J$ has the property $\|J\| = \sup_{\|x\|_v} \|Tx\|_v = |\lambda|$.
I think the answer is no. We know any vector induced norm has the property $\lambda \le \|J\| \le \lambda + \varepsilon$ for every $\epsilon > 0$. But in the classical construction for this proof, it seems like we cannot reduce $\epsilon$. But how to prove this rigorously? Suppose we have some vector norm on $\mathbb C^2$. Let $x \in \mathbb C^2$ with $\|x\|_v = 1$. How do we claim following strict inequality \begin{align*} \|J\| = \sup_{\|x\|_v=1} \| Jx \|_v > |\lambda|. \end{align*}