# Perturbation parameters in Eigenvalue question [easy]

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?

-

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.
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