Fastest way of finding eigenvectors from eigenvalues Given the eigenvalue of a matrix of large dimensions, I want to know if there is a fast way of finding the corresponded eigenvectors?
 A: I don't know if it's the fastest way but recently Denton et al. showed that (https://arxiv.org/abs/1908.03795, Lemma 2)
$$
|v_{i,j}|^2\prod_{k=1;k\ne i}^n(\lambda_i(A)-\lambda_k(A))=\prod_{k=1}^{n-1}(\lambda_i(A)-\lambda_k(M_j))
$$
where 


*

*A is an $n \times n$ Hermitian matrix with eigenvalues $\lambda_i(A)$ and normed eigenvectors $v_i$,

*and the elements of each eigenvector are $v_{i,j}$,

*and $M_j$ is the $n − 1 \times n − 1$ submatrix of A that results from deleting the jth column and the jth row, with eigenvalues $\lambda_k(M_j)$. 

A: Please refer to Finding Eigenvectors: Fast & Nontraditional way or the arXiv preprint for fast and Nontraditional approach without using the Gaussian-Jordan elimination process.  
When the matrix is diagonalizable (There is a way to check that) and has a spectrum of two, there is no need to evaluate eigenvectors at all since they already appear as nonzero column vectors of certain matrices that we would like to call The eigenmatrix. We have given a general theory for diagonalizable and nondiagonalizable matrices as well. 
You have access to the part of the preprint under the same link. 
