I know that we can find the largest eigenvalue of a matrix by taking a "random" vector $v$ and then calculating $v_k:=A^k$v. Defining: $\lambda_k :=\frac{v_k^Tv_{k+1}}{\|v_k\|_2^2}$ we have $|\lambda_k-\lambda_{max}|= \mathcal{O}(|\frac{\lambda_2}{\lambda_1}|^k)$, with $\lambda_{max}$ the largest eigenvalue by absolute value.
This can be generalised to find any eigenvalue $\lambda$ because if we have an estimate $\mu \approx \lambda$ we can apply above procedure to $(A-\mu \mathbb{I})^{-1}$ which should have larges eigenvalue $\lambda$. Im wondering under which condition this inverse is well defined (provided A is invertible): Since $(A-\mu \mathbb{I})=(\mathbb{I}-\mu A^{-1})A$ we can show that $(\mathbb{I}-\mu A^{-1})$ is invertible, by the Neumann series we get: $$ (\mathbb{I}-\mu A^{-1})^{-1}=\sum_{j=0}^{\infty}\mu^{j}A^{-j} \Rightarrow \|\sum_{j=0}^{\infty}\mu^{j}A^{-j}\| \leq \sum_{j=0}^{\infty}|\mu|^j \|A^{-1}\|^{j} $$ with a submultiplicative norm $\|\cdot \|$. This only converges if $|\mu|\|A^{-1}\|<1$. Hence we dont seem to be free to choose $\mu$, it needs to be small, which doesnt make sense. I'd be happy if someone knows any better results regarding the existance of this inverse.
The inverse exists exactly when $\mu$ is not an eigenvalue. This is practically equivalent to the definition of an eigenvalue, $λ$ is one if and only if $A-λI$ has a non-trivial kernel, i.e., is singular.
If $\mu\approx λ$ is not an eigenvalue but close to one and only one, then $(A-\mu I)^{-1}$ has $(λ-\mu)^{-1}$ as largest eigenvalue.
The method is usually called the "shifted inverse power iteration".