Tricks for rapidly calculation of $e^A$

Asked Oct 12 '17 at 03:09

Active Oct 12 '17 at 13:50

Viewed 87 times

I am looking for tricks for rapidly calculation of $e^A$. It is not a problem to assume the dimension of the matrix is small ($\leq3$).

For example: if $A$ is a 2x2 diagonalizable matrix and $\lambda,\mu$ are their eigenvalues, then $A=P^{-1}DP$, where $D$ is a diagonal matrix with $\lambda,\mu$ as entries. Then $A^n=P^{-1}D^nP$ for some invertible matrix $P$ and, therefore, $A^n_{ij}=\alpha_{ij}\lambda^n+\beta_{ij}\mu^n$ for all $n$. The constants can be easily calculated just with $A^0$ and $A$ and a value for $e^A$ follows easily from this.

edited Oct 12 '17 at 13:50

asked Oct 12 '17 at 03:09

Veridian Dynamics

2,521

6

Your post does not contain a question. – symplectomorphic Oct 12 '17 at 03:15
2

No need to find an eigenbasis. Once you have the eigenvalues, you can decompose diagonalizable $A$ into a linear combination of projections, which have simple exponentials. See this answer for all of the possibilities for a $2\times2$ matrix. – amd Oct 12 '17 at 03:52
1

@symplectomorphic I am asking for "tricks for rapidly calculation o $e^A$" – Veridian Dynamics Oct 12 '17 at 13:46
@amd Thanks, your link helped me a lot. – Veridian Dynamics Oct 13 '17 at 00:54

Tricks for rapidly calculation of $e^A$

0 Answers0