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Consider the matrix $A = \begin{pmatrix} 2 & -3 \\ 1 & 4\end{pmatrix}$, which should represent a system of linear differential equations. I need to find the flow $\varphi(t, X)$ whereas $X$ is the initial value. Thus I need to calculate the exponential matrix $e^{tA}$.

In order to do that I determined the eigenvalues $\lambda_{1,2} = 3 \pm i \sqrt{2}$. Here is where I encounter some problems. I think I could determine the exponential matrix by saying: $A = SDS^{-1}$, whereas $D$ is the corresponding rotation matrix and $S$ consists of the (complex) eigenvectors of $A$ (and than using some characteristics of the exponential matrix).

However, I found out that apparently one can write $A$ as $A = D + N$ where $D$ is diagonalizable and $N$ is nilpotent (and $DN = ND$), since the characterstic polynom of $A$ can be written as linear factors. This approach seems to be easier than the solution I have noted above - depending on how easy it is to determine $D$ and $N$. Thus, my question: How do I determine $D$ and $N$? Is there a standard method or is it just "seeing"/guessing how you could obtain those two matrices?

Thanks in advance for your help!

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Does the decomposition $A=D+N$ have a name? – Git Gud Feb 26 '13 at 20:39
@GitGud I'm not sure. On Wikipedia I found that if the characteristic polynom can be written as linear factors (which is always the case in $\mathbb C$), than such a decomposition exists. They didn't mention a name though. – StringerBell Feb 26 '13 at 20:57
How can I find it. Are you wanting to finding this decomposition or would an alternative be good as well? – Git Gud Feb 26 '13 at 20:59
@GitGud See here, point "Generalization". Actually, I was wondering how to find this decomposition. But, of course if you know of an alternative (which is not the one I've already mentioned), I'd be grateful as well. – StringerBell Feb 26 '13 at 21:05
I don't think there's a quick answer to what you're looking for. You probably gotta go with diagonalizability. Even JNF wouldn't save you from complex eigenvalues. I don't think there's a way around that. – Git Gud Feb 26 '13 at 21:16
up vote 1 down vote accepted

Note that since $N$ is nilpotent, its only eigenvalue is $0$. If you were to write $A$ as $A=D+N$, you'd have $D=A-N$ which would imply that the eigenvalues of $D$ are those of $A$. So when you diagonalize $D$ you'll get complex eigenvalues again, so might as well use $A$'s diagonalizability from the get go. I don't think there's a way around the complex eigenvalues.

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Since you already know that alleigenvalues of $A$ are distinct (i.e. there is no multiple eigenvalue, the matrix $A$ itself is diagonalizable, i.e. $D=A, N=0$. Go ahead with your first ideas, actually diagonalizing $A$ (see? It's diagonalizable).

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"Thus, my question: How do I determine $D$ and $N$?" – Git Gud Feb 26 '13 at 20:15
But $A$ has complex eigenvalues, so I think $A$ is not diagonalizable, at least not in $\mathbb R$, right? Thus I couldn't say $D = A$. I was actually just wondering how it is possible to decompose $A$ into $D$ and $N$ ... – StringerBell Feb 26 '13 at 21:01

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