2-norm of a canonical Jordan form and spectral radius Let $J$ be a canonical Jordan form (real or complex). Is it true that the $2$-norm of $J$ is equal to its spectral radius?
 A: The answer is no. Consider the following Jordan cell
$$
A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.
$$
Eigenvalues are $\lambda_1=\lambda_2=1,$ spectral radius is $\rho(A)=\max_{k=1,2}|\lambda_k|=1,$ spectral norm is
$$
\|A\|_2 = \max\frac{|Ax|_2}{|x|_2} \geq \frac{\left| A e \right|_2}{|e|_2} = \frac{\left|\begin{bmatrix}2 \\ 1\end{bmatrix}\right|_2}{\left|\begin{bmatrix}1 \\ 1\end{bmatrix}\right|_2} = \frac{\sqrt{5}}{\sqrt{2}}=\sqrt{2.5}\approx 1.58,
$$
where $e=\begin{bmatrix}1 \\ 1\end{bmatrix}.$ 
The actual norm is even larger, try to find it yourself.
A: Assume that $J=\bigoplus_{i=1}^kJ_i$, where $J_i$ is an elementary Jordan block of the dimension $n_i$ ($\sum_{i=1}^k=n$) corresponding to an eigenvalue $\lambda_i\in\mathbb{C}$. Then
$$
\rho(J)=\max_{i=1,\ldots,n}|\lambda_i|, \quad \text{and}\quad
\|J\|_2=\max_{i=1,\ldots,n}\|J_i\|_2,
$$
so the insight related to whether or not this is true can be obtained by examining the validity of the statement for elementary Jordan blocks.
For an elementary Jordan block $J_i\in\mathbb{C}^{n_i\times n_i}$ corresponding to an eigenvalue $\lambda_i\in\mathbb{C}$, we have:

$$\rho(J_i)=\|J_i\|_2 \quad\Leftrightarrow\quad n_i=1.$$

Proof: It is easy to see that $\rho(J_i)=\|J_i\|_2$ if $n_i=1$.
Now assume that $n_i\geq 2$.
Consider the diagonal entry $(n_i,n_i)$ of the matrix $J_i^*J_i$, which is equal to $|\lambda_i|^2+1$. Since the diagonal entries of an HPD matrix bound from below its spectral radius, it follows that
$$
\|J_i\|_2^2=\rho(J_i^*J_i)\geq|\lambda_i|^2+1>|\lambda_i|^2=\rho^2(J_i).
$$
Some sufficient conditions for $\rho(J)=\|J\|_2$:


*

*Obviously, if $J$ is diagonal then $\rho(J)=\|J\|_2$.

*From the Gershgorin theorem applied on $J_i^*J_i$, one can see that
$$\tag{$*$}
|\lambda_i|\leq\|J_i\|_2\leq|\lambda_i|+1.
$$
Assume that $\rho(J)=|\lambda_j|$ for some $1\leq j\leq k$ and let $n_j=1$. If $|\lambda_i|+1\leq|\lambda_j|=\rho(J)$ for all $i\neq j$ (this is a sufficient condition for $\|J_i\|_2\leq|\lambda_j|$), then $\rho(J)=\|J\|_2$. Actually, the condition that $\|J_i\|_2\leq|\lambda_j|$ for all $i\neq j$ might be also a necessary condition.
Note that the bounds in ($*$) are sharp. The lower bound is attained for $n_i=1$ while the upper bound is attained asymptotically for $n_i\rightarrow\infty$.
