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Is there a transformation to decrease the dimension of a matrix so that it becomes full rank?

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What are you actually trying to do? –  J. M. May 12 '12 at 15:50
    
I'm playing around with power methods for solving the eigen problem. After I solve for the first eigenvector and set up the system for solving the next eigenvector I notice that the converges is much slower (empirical result), I'm wondering if that is because the matrix is no longer full rank. –  csta May 12 '12 at 16:03
    
Well, the power method was always intended for getting just the dominant eigenpair. If you need to pick eigenpairs, look up the inverse power method. –  J. M. May 12 '12 at 16:13
    
The inverse power method is even slower then the traditional power method, as my system is fairly large (~$10^4$). Also I believe my question is well posed in the sense of stack.exchange. –  csta May 12 '12 at 17:23
    
... your having a large matrix seems to have not been mentioned in your OP. Is this large matrix of yours completely and utterly random, or did it come from somewhere, and thus has attendant structure (e.g. symmetry, sparsity) to it? –  J. M. May 12 '12 at 17:29

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