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I'm solving an eigenvalue/eigenvector question of the matrix:

\begin{bmatrix} 2 & 1 \\ 0 & 2 + \varepsilon \end{bmatrix} where $\varepsilon$ is the perturbation parameter. Would I just solve this as though $\varepsilon$ were a variable? Or does it have a special property?

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It has a special property that one of the elements is zero. The characteristic equation is: $(\lambda - 2)(\lambda - (2+\epsilon))=0$ and therefore eigenvalues are simply the elements of the main diagonal.

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I'm asking about a special property of $\varepsilon$. I know how to compute eigenvalues/eigenvectors, but in doing so should I treat the perturbation parameter just as I would treat any variable? –  David Apr 30 '13 at 9:13
    
It is treated as a constant when finding the eigenvalue. –  genepeer Apr 30 '13 at 9:20
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