calculating eigenvector components of a given vector

Question

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?

Answer 1

Since the matrix is hermitian, you could use it as a hamiltonian to propagate it in imaginary time. That is, solve the following system of differential equations:

$$ i\frac{d \vec{V}}{dt}=M\vec{V}$$

The general solution to this is:

$$ V(t)=V_0e^{iMt}$$

Then you take your $\vec{V(t)} \cdot \vec{V(0)}$, fourier transform it, and the height and placement of the peaks will tell you the components along various eigenvectors and their associated eigenvalues. This is sometimes called "the spectral method" in ultrafast atomic physics.

Once you have the eigenvalues, find the eigenvectors with whatever specific-eigenvalue solver you prefer.

Answer 2

Sure. Let $\hat{M}$ be a matrix such that each column is a distinct eigenvector of the Hermitian operator $M$; it follows from the Hermiticity of $M$ that $\hat{M}$ is invertible. To obtain the $v_{i}$, solve the linear system $\hat{M}v = V$. Then, select the index $i$ such that $\max_{j}|v_{j}| = |v_{i}|$; the $i$th column of $\hat{M}$ will the eigenvector corresponding to the largest $v_{i}$ in magnitude.