Non-diagonalizable 2x2 matrix norm strictly greater than spectral radius

Question

I've seen many answers to problems similar to this, yet I couldn't solve it.

Let $J= \mathbb{C}^{2x2}$ be the matrix: $$J = \begin{pmatrix}\lambda & 1\\ 0 & \lambda\end{pmatrix}$$

Show that for every norm of $\mathbb{C}^2$, $||J|| = \max_\limits{||x|| = 1} ||J\,x|| > |\lambda|$.

I also have the following hint.

Hint: Calculate $||J^k\, x||$, where $x = (1\ \lambda)^T$.

Proving $||J|| \geq |\lambda|$ is easy, and I was able to show that $||J^k\, x|| = |\lambda|^k ||k\ v + x||$, where $v = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is the eigenvector associated with $\lambda$. I'm also aware that $v$ can be freely chosen so that $||v|| = 1$.

However, I haven't been able to prove that it is strictly greater than $\lambda$.

Answer 1

A proof that doesn't require us to know anything about the underlying norm on $\Bbb C^2$:

Wlog $\lambda\ne0$. Let $$A=\begin{bmatrix}1&1/\lambda\\0&1\end{bmatrix}.$$We need to show that $$||A||>1.$$

Lemma. Suppose $E$ is a $2\times 2$ matrix, $E\ne0$ and $E^2=0$. Then $||I+E||>1$.

Proof: $(I+E)^n=I+nE$ for $n=1,2,\dots$, so $$||I+E||^n\ge||(I+E)^n||\ge n||E||-||I||\to\infty.$$