Find perturbed eigenvalues, eigenvectors by perturbation methods Suppose we know $A_0v_0^k = \lambda_0^k v_0^k$, we can find the following perturbed eigenvalues and eigenvectors $$(A_0+\epsilon A_1)v(\epsilon) = \lambda(\epsilon) v(\epsilon)$$ by assuming that $$\lambda(\epsilon)=\sum_{j=0}^\infty \lambda_j\epsilon^j$$ and $$v(\epsilon)=\sum_{j=0}^\infty v_j\epsilon^j$$ and then plug in to find the coefficients. However, suppose I have the following matrix: $$M = \begin{bmatrix}a & b\\\ \frac{c}{\epsilon} & \frac{d}{\epsilon}\end{bmatrix}$$ Now, obviously, if I let $\epsilon=0$, then both entries will go to infinity (singular). So how do I modify this such that I can still use the perturbation method to find the perturbed eigenvalues and eigenvectors? I just think about the following way $$M = \frac{1}{\epsilon} \begin{bmatrix} \epsilon a &\epsilon b\\\ c & d\end{bmatrix}$$ but I have no idea how to go further. Please advise, thanks!

As your matrix is singular in $\epsilon$, it is not too far-fetched to suspect that its eigenvalues will be singular, too. This is supported by the relation between the trace and determinant of a matrix and its eigenvalues, see e.g. here#Eigenvalue_relationships). So (ignoring the fact that for a $2\times 2$ matrix you can calculate its eigenvalues explicitly, and expand the result in powers of $\epsilon$), it seems a good idea to use the series $$ \lambda(\epsilon) = \sum_{j=-N}^\infty \lambda_j \epsilon^j, $$ for some suitable choice of $N$. In your case, $N=1$ will suffice; of course, you can also choose $N=2$ (or higher), but this will just yield $\lambda_{-2} = 0$. Using this approach, you'll find $$ \lambda_1 = \frac{a d- b c}{d} + \mathcal{O}(\epsilon^1)\quad\text{and}\quad \lambda_2 = \frac{d}{\epsilon} + \mathcal{O}(\epsilon^0). $$