# Representation of Jordan block for exponent matrix

Given $A = n\times n$ matrix with the real parts of its eigenvalues are contained in $[\alpha, \beta]$ where $-\infty < \alpha \leq \beta <\infty$. I'm currently trying to find a way to decompose $A$ into Jordan block such that $e^{At}$ can be rewritten as the exponent of linear combination of Jordan blocks + some off-diagonal matrices. For example, if $A$ has a single Jordan block corresponding to eigenvalue $\lambda$, then $A = \lambda I + B$, where $B =$ an off-diagonal nilpotent matrix. I'm trying to generalize this special result to any numbers of Jordan block that $A$ might have, but I still fail:( Can anyone please help me with this problem?

• What have you tried? Have you played with some specific example matrices? Also note that the inequalities on the eigenvalues are irrelevant to the question. – user7530 Sep 30 '15 at 6:12
• @user7530: I tried doing Jordan canonical form for $A$, but when I followed the normal process even with specific matrices, all I got is the 'useless' transformation $T^-1AT = J$ where $J$ is Jordan matrix. The reason I said it's useless is because I want to express $A$ as the sum of certain matrices, so that I can use Taylor's expanson for $e^{\text{that sum}\times t}$, as your method shows. When I got that, it's done!! Currently, I don't see how to decompose $A$ into that summation form so that I can bound the norm of the terms $t^{k}B^k$:P – user177196 Sep 30 '15 at 18:56

Jordan-decompose $A$ as $A=U^{-1} J U$ where $J$ is block-diagonal with blocks $J_i$. Let $K_i$ be the matrix consisting of $J$ with all blocks except for $J_i$ zeroed out. Then $J=\sum K_i$ and clearly $K_iK_j = K_jK_i$, so
$$e^{At} = U^{-1} \left( \prod e^{K_i t}\right) U.$$