# 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).)

• Which matrix norm are you using? Apr 4, 2016 at 5:20
• No. Take the Frobenius norm. Then $\|J\|^2 = \sum_i \|J_j\|^2$. Apr 4, 2016 at 5:33
• Sorry I should have clarified. I used Matlab for my examples. The norm would be the one that Matlab uses when I type norm(A). Apr 4, 2016 at 5:44
• That's the induced 2-norm, i.e. the largest singular value. Apr 4, 2016 at 5:49

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.