I have some vector $V$ which can be decomposed into the eigenspace of the hermitian sparse operator $M$:$V = \sum_i v_i \hat{m}_i$Is there a way to find the $\hat{m}_i$ (the eigenvector itself) that correspond to the largest $v_i$ (in magnitude)?I essentially want the largest few terms of the sum, including the eigenvectors of $M$, which I don't know ahead of time.Specifically, I want to simultaneously find the eigenvectors of $M$ that correspond to the largest $|v_i|$, along with finding the largest $v_i$.  Preferably without finding the entire spectra of $M$ first.Some possibilities that I have been thinking about:We can inflate the matrix using the opposite of Wieldant's Deflation:$M_1 = M + \sigma \left[ \Sigma_i v_i \hat{m}_i \right] V^H  = M + \sigma V V^H$The eigenvalues for different $\hat{m}_i$ are shifted $\lambda_i + \sigma |v_i|^2$.  I believe we can then extract $\sigma$ and $v_i$ because the eigenvectors don't change.  The problem is that the outer product of $V$ is dense.another possibility:The power method (keep multiplying $M$ by our vector $V$ until the convergence) finds the component of $V$ with the largest eigenvalue.  The downside of this method is that we don't control for the magnitude of $v_i$, so we would end up finding ALL the components, and then finding the largest.Is there some way to control this so that we only converge on the largest component?
calculating eigenvector components of a given vector
sparse;eigensystem