With Power Method, I can find the largest eigenvalue and the corresponding eigenvector. But what if I have and $n,n$ matrix e.g $4,4$
$$A = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 3 \end{bmatrix}$$
And then I want to compute the eigenvalues. $$|A - \lambda I| = \begin{vmatrix} 2 - \lambda & 0 & 0 & 0 \\ 1 & 2- \lambda & 0 & 0 \\ 0 & 1 & 3- \lambda & 0 \\ 0 & 0 & 1 & 3- \lambda \end{vmatrix} = (2 - \lambda)^2(3 - \lambda)^2$$
Here it's easy to find that the eigenvalues are 2 and 3. But with the power method, I would only get number 3. Is a way to re-compute the power method with that eigenvalue 3 and find eigenvalue 2?