Relation between the subordinate norm and the spectral radius of a matrix

Question

Let's define the following subordinate norm of a $(NM \times NM)$ matrix A norm as follows

\begin{eqnarray*} ||A||_{2,b} = \mathrm{max}_{x \in \mathbb{C}^{NM}} \left \{ \frac{||A x||_b}{||x||_2} \right \} \end{eqnarray*}

where $||.||_2$ denotes the Euclidean norm of the $(NM \times 1)$ vector x, and where $||x||_b$ denotes the block maximum norm of the $(NM \times 1)$ vector x, i.e.

\begin{eqnarray*} ||x||_{2,b} = \mathrm{max}_{1 \leq k \leq N} \left \{ ||x_k||_2 \right \} \end{eqnarray*}

with $x = [x_1^T \, \, x_2^T \ldots \, x_N^T]$ and $x_k$ equal to a $(M \times 1)$ sub-vector of x for all $k$. It is well-known that

$||A||_2 \geq \rho(A)$

and that

$||A||_b \geq \rho(A)$

where $\rho(A)$ denotes the spectral radius of A (i.e. $\mathrm{max} |\lambda_i|$ with $\{\lambda_i\}$ denoting the eigenvalues of the matrix $A$), and where \begin{eqnarray*} ||A||_{2} = \mathrm{max}_{x \in \mathbb{C}^{NM}} \left \{ \frac{||A x||_2}{||x||_2} \right \} \end{eqnarray*} and \begin{eqnarray*} ||A||_{b} = \mathrm{max}_{x \in \mathbb{C}^{NM}} \left \{ \frac{||A x||_b}{||x||_b} \right \}. \end{eqnarray*} However, I would like to know if $||A||_{2,b} \geq \rho(A)$. By using the material that there is in the literature, I am only able to prove that $||A||_{2,b} \geq \rho(A)c$ with $c \leq 1$.

Answer 1

This ratio $\|A\|_{2,b}/\rho(A)$ can be arbitrarily small. Consider the $N\times N$ matrix $A$ all of whose entries are equal to $1$, and interpret this as a block matrix with blocks of size $M=1$. Then $\rho(A)=N$, $$ \|Ax\|_b = \max_j |x_1+\ldots + x_N| \le \|x\|_1 , $$ so $\|A\|_{2,b}\le \sup \|x\|_1/\|x\|_2 = \sqrt{N}$.