My Math Forum Positive definite matrix whose smallest eigenvector is predefined

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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).

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