Is the matrix norm of a matrix equal to the maximum of the norms of its Jordan block? Let $J$ be a Jordan block matrix with blocks $J_1,\cdots,J_n$. I came up with some examples of $J$ and noticed that $\|J\|=\max_{i=1,\cdots,n}\|J_i\|$. Does this result always hold?
The norm I use here is the induced 2-norm. (I used Matlab for my examples. The norm would be the one that Matlab uses when I type norm(A).) 
 A: The answer is yes.  Let $$J = \left( {\begin{array}{*{20}c}
   \hfill {J_1} & \hfill {} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {J_2} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {} & \hfill  \ddots  & \hfill {} \\
   \hfill {} & \hfill {} & \hfill {} & \hfill {J_n} \\
\end{array}} \right)$$ be your Jordan block matrix.  Each Jordan block $J_i$ has a singular value decomposition $J_i  = U_i \Sigma _i V_i ^* $, so we can write out a singular value decomposition for $J$ as $$J = \left( {\begin{array}{*{20}c}  \hfill {U_1} & \hfill {} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {U_2} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {} & \hfill  \ddots  & \hfill {} \\
   \hfill {} & \hfill {} & \hfill {} & \hfill {U_n} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   \hfill {\Sigma _1} & \hfill {} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {\Sigma _2} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {} & \hfill  \ddots  & \hfill {} \\
   \hfill {} & \hfill {} & \hfill {} & \hfill {\Sigma _n} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   \hfill {V_1^* } & \hfill {} & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {V_2^* } & \hfill {} & \hfill {} \\
   \hfill {} & \hfill {} & \hfill  \ddots  & \hfill {} \\
   \hfill {} & \hfill {} & \hfill {} & \hfill {V_n^* } \\
\end{array}} \right).$$
The fact that we can express the singular value decomposition of $J$ in compatible block form with the Jordan blocks $J_i$ is essentially the only observation needed to answer your question.  
$\left\| J \right\|_2  = \sigma _{\max}(J)$, so assume $\sigma_{max}(J)$ belongs to $\Sigma_k$ (it has to belong to some $\Sigma_i$ for $1\le i\le n$).  Then, since $J_k=U_k\Sigma_k V_k^*$, we have that $$\|J\|_2=\sigma_{max}(J)=\|\Sigma_k\|_2=\|J_k\|_2=\max_{i=1,\cdots,n}\|J_i\|_2.$$
Note that this argument does not involve anything specific about the Jordan block structure.  It remains valid provided $J$ has any diagonal block structure.
