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Given the Matrix, $ A:=\begin{bmatrix}6 & 9 &15 \\ -5&-10 & -21 \\2&5&11\end{bmatrix} $, i have worked out the JordanNormalForm, J:= $\begin{bmatrix}3 & 0 &0 \\ 0&2 & 1 \\0&0&2\end{bmatrix} $ and the Transition Matrix, P:= $\begin{bmatrix}1 &-3 &-3 \\-2 &3 &1 \\1&-1&0\end{bmatrix} $

However, the question i'm stuck on requires me to "use the Jordan normal form J and the transition matrix P for computing $e^{At}$ in exact form"

I know $e^{At} = P.e^{Jt}.P^{-1}$ but im not sure how to go about computing $e^{Jt}$??

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  • $\begingroup$ This is of help. $\endgroup$ – Git Gud Mar 2 '13 at 22:17
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Using this (or verifying manually) we can see that $$\pmatrix{2&1\\0&2}^n=\pmatrix{2^n&n\cdot 2^{n-1}\\0&2^n} =2^n\cdot \pmatrix{1&n/2\\0&1},$$ So, for this Jordan block, $$e^{\ \ t\cdot\pmatrix{2&1\\0&2}} = \sum_{n\ge 0} \frac{(2t)^n}{n!}\pmatrix{1&n/2\\0&1} = \\ = \pmatrix{e^{2t} & t\cdot e^{2t} \\ 0 & e^{2t} } $$ calculating the sum of the series componentwise in the matrix, and using $$\sum_{n\ge 0}\frac{(2t)^n}{n!}\cdot\frac n2 = \sum_{n\ge 1}t\cdot\frac{(2t)^{n-1}}{(n-1)!}\ .$$

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