Dominant eigenvectors of an unknown matrix

Question

Do you have any idea about how we can find the principal eigenvectors of an unknown matrix ${H}$?

The elements of $H$ are unknown in general. If you are familiar with channel estimation procedure in wireless communications, $H$ is channel matrix, and every element $(i,j)$ represents the channel gain between antenna element $i$ of transmitter and antenna element $j$ of the receiver. The following are the information that we have:

1. we do not know the channel matrix,
2. we know second order statistics, covariance matrix of the channel,
3. the channel matrix size tends to infinity (large number of antennas in Tx and Rx),
4. the channel is sparse, that is, it has only two or three dominant eigenvectors regardless of the channel matrix size.

Note that the power method is not a good solution for me, since its complexity is huge, especially because it requires many iterations to find the dominant eigenvectors.

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Some information about the interplay between channel and its covariance:

We consider Rayleigh correlated channel coefficients $h_k\sim CN (0,\mathbf{R}_k)$, mutually independent across the users $k$, and $CN$ is circularly symmetric complex Gaussian. We let $\mathbf{R}_k = \mathbf{U}_k \mathbf{\Lambda}_k \mathbf{U}_{k}^{H}$, where $\mathbf{U}_{k}$ is a tall matrix of the eigenvectors corresponding to the non-zero eigenvalues of $\mathbf{R}_k$, given as the diagonal elements of the diagonal matrix $\mathbf{\Lambda}_{k}$.

The Karhunen-Loeve representation of the user channel vector is given as $\mathbf{h}_{k} = \mathbf{U}_{k}\mathbf{\Lambda}_{k}^{\frac{1}{2}}\mathbf{w}_{k}$, where the entries of $\mathbf{w}_{k}$ are i.i.d. with zero mean Gaussian distributed with variance 1.

Answer 1

In the absence of other answers, and sorry if you have already considered this, but the power method can be very fast with a sparse matrix. You may like to check out how Google does it (eigenvectors are used for page rank, which is pretty fast).

http://en.m.wikipedia.org/wiki/PageRank

Because of the large eigengap of the modified adjacency matrix above,[22] the values of the PageRank eigenvector can be approximated to within a high degree of accuracy within only a few iterations.

Footnote 22 points to: http: //www-cs-students.stanford.edu/~taherh/papers/secondeigenvalue.pdf

Footnote 23 discusses how to make the calculation even faster http: //projecteuclid.org/euclid.im/1150474883

I have not read the papers, but I hope that you find one of them to be useful. (I had to break the links as new users can not post many links.)