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 September 21st, 2015, 11:45 PM #1 Newbie   Joined: Oct 2014 From: China Posts: 12 Thanks: 2 Positive definite matrix whose smallest eigenvector is prescribed I have two twisted questions: 1) A is a N by N positive definite symmetric matrix, thus $AP=P \Omega$, the columns of P are eigenvectors; $\Omega$ is the diagonal matrix consist of the eigenvalues. Knoeppel’s 2013 paper says the smallest eigenvector $\psi$ could be parametrically aligned to any vector $\phi$ by $$(A-tI)\psi = \phi$$ When t equals to the smallest eigenvalue, there is no alignment; when t approaches to $-\infty$ the alignment is stronger. Now, My goal is to align the $i$th (instead of the smallest) eigenvector $\nu$ to $\phi$. 2) Thus I come up with a ‘solution’: construct a new positive definite matrix B whose smallest eigenvector equals to the $i$th eigenvector $\nu$ of A, then constructing B via $$B=P \Omega P^{-1}$$ wherein P consists of the newly constructed eigenvectors of B, finally $(B-tI)\psi = \phi$ would work for my purpose. But I doubt whether it’s possible. Suppose I want the smallest eigenvector of B equals to the 2nd (smallest) eigenvector of A. I can shift $i$th eigenvector of A as the $(i-1)$th eigenvector of B, and adding a new eigenvector $\eta$ to B. Whereas, the contradiction is: the $N-1$ eigenvectors (excludes the smallest ) of A already determinate the $\eta$ since all eigenvectors are pair-wise orthogonal. Another strategy might be keeping the $i$th eigenvector $\nu$ of A as the smallest eigenvector of B, and invent all $N-1$ eigenvectors of B. It is equivalent to find an arbitrary set of orthogonal vectors in the $N-1$ dimensional subspace. However, how to make sure that the eigenvalue of $\nu$ is smaller than the eignvaues of all newly constructed eigenvectors? Sorry for a long question. Last edited by whitegreen; September 21st, 2015 at 11:51 PM. September 25th, 2015, 05:41 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The "smallest eigenvector" makes no sense because any multiple of an eigenvector is an eigenvector. I think you mean "an eigenvector corresponding to the smallest eigenvalue (in absolute value). Tags definite, eigenvalue, eigenvector, matrix, positive, predefined, smallest Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Alann Number Theory 7 November 7th, 2012 08:39 AM Linear Algebra 2 January 19th, 2012 04:08 AM billybob Calculus 5 May 12th, 2011 02:37 AM krystoferivanov Linear Algebra 0 December 19th, 2010 03:49 AM skeboy Linear Algebra 0 June 11th, 2009 10:51 PM

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