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?

  • $\begingroup$ 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. $\endgroup$ – user7530 Sep 30 '15 at 6:12
  • $\begingroup$ @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 $\endgroup$ – 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

$$e^{At} = U^{-1} \left( \prod e^{K_i t}\right) U.$$

Now bound each term of the product using the other answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.