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Given the matrix: $A=\begin{pmatrix}-2&0&0\\4&-2&0\\1&0&-2\end{pmatrix}$, find $e^{At}$.

I found the eigenvalues to be $-2,-2,-2$. I need to use the Jordan form to solve it. I'm practicing for exam in a couple of hrs. Thank you. I found the eigenvectors, $$v_1=(0,1,0), v_2=(0,1,1), V_3=(1,0,0).$$

I think the Jordan form is $e^{Jt}=\begin{pmatrix}e^{-2t}&1&0\\0&e^{-2t}&1\\0&0&e^{-2t}\end{pmatrix}$,Is this right?

Next I need to calculate $$Ve^{Jt}V^{-1}$$ Actually I do know the answer Im just practicing the way to do it. Can anybody tell me whats wrong with V_3? I found that $(A+2I)^3=0$ and I thought any eigenvector satisfies $0\cdot v_3=0$?!

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    You'll need the eigenvectors for the next step.2012-11-28
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    @ hans what is the form of E^Jt?2012-11-28
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    $V_3$ is not an eigenvector.2012-11-28
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    @Hans how do you find the second generalized eigenvector? how can you tell that it is not?2012-11-28
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    Here is another hint; see the worked example under calculations (assuming you understand Hans' comment about the need for the eigenvectors): http://en.wikipedia.org/wiki/Matrix_exponential2012-11-28
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    @ amzoti how do you find the second generalized eigenvector?2012-11-28
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    See the examples here (do you understand algebraic and geometric multiplicity): http://www.math.vt.edu/people/afkhamis/class_home/notes/F08W12.pdf2012-11-28
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    @Klara - use the eigenvectors $e_2 = (0,1,0)^T$ and $e_3 = (0,0,1)^T$ for the single eigenvalue $\lambda = -2$. The missing generalized eigenvector $x$ has the property that $(A - \lambda I) x = \alpha e_2 + \beta e_3$. When you write out what this means, you'll see that $x = (1,0,0)^T$ works. Good luck with your exam.2012-11-28

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