How to derive the Euclidean norm of a matrix from its spectral radius [closed]

Question

For a matrix $A\in R^{n \times n}$, if there is a positive $\rho \in (0,1)$ s.t. $\rho(A) \leq \rho$, where $\rho(A)$ represents the spectral radius of $A$.

Can we get the following conclusions:$\Vert A \Vert_2 \leq \rho$ ?

For example, let \begin{equation} A=\left[ \begin{array} &1-3w& h &1 \\ -3w^2 &1 &h\\ -w^3& 0& 1 \end{array} \right]. \end{equation} Let $hw \in (0,1)$, it is obvioulys that the spectral radius of $A$ is $1-hw$. If there is a positive $\rho \in (0,1)$ s.t. $1-hw \leq \rho$, can we obtain a upper bound less than 1 of $\Vert A \Vert_2$ ?

Answer 1

The correct statement is known as Householder Theorem: For every $\epsilon>0$, there exists a norm $\|\cdot\|$ over ${\mathbb C}^n$, such that the subordinated norm of $A$ is $\le\rho(A)+\epsilon$.

As mentionned by Fedor, the result is false with $\epsilon=0$ because if $A\ne0_n$ is nilpotent, then $\rho(A)=0$ and $\|A\|=0$ for every norm.