MT 김신휘 Computation of e^A using Jordan Canonical Form

MT Page 334 Problem 9.4

$A=\begin{bmatrix} 2 & 1 & 0 & 0\\ 0 &2 & 1 &0 \\ 0 &0 &2 & 0\\ 0 &0 & 0 &1 \end{bmatrix}$

(2) Computation of the exponential matrix $e^{A}$ of $A$.

(Citation : Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 334p)

Sol)

Let $J$ be a Jordan form of $A$ such that $A=QJQ^{-1}$.

.

Also, when $J_{i}=\lambda I+N$,

$e^{J_{i}}=e^{\lambda I}e^{N}=e^{\lambda }\sum_{k=0}^{n-1}\frac{1}{k!}N^{k}=e^{\lambda }\begin{bmatrix} 1 & \frac{1}{1!} & \frac{1}{2!} &\cdots &\frac{1}{(n-1)!} \\ 0 &1 & \frac{1}{1!} &\cdots &\frac{1}{(n-2)!} \\ 0& 0 &1 & \cdots &\frac{1}{(n-3)!} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 0 & 0 & 0 &\cdots &1 \end{bmatrix}$.

이제 sage를 이용해서 $e^{A}$를 구하자.

(Citation : Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 334-335p)

Note

- $e^{QJQ^{-1}}=Qe^{J}Q^{-1},\: \: e^{\lambda I}=e^{\lambda }$

- $e^{A}$를 sage로 직접 구하는 것은 워크시트에서는 되지 않지만 공개를 한 문서에서는 가능하다.

Reference

Jin Ho Kwak and Sungpyo Hong, 1997, Linear Algebra, 334-335p

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