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September 22nd, 2015, 12:45 AM  #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 $$(AtI)\psi = \phi$$ 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 $(BtI)\psi = \phi$ 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 $(i1)$th eigenvector of B, and adding a new eigenvector $\eta$ to B. Whereas, the contradiction is: the $N1$ eigenvectors (excludes the smallest ) of A already determinate the $\eta$ since all eigenvectors are pairwise orthogonal. Another strategy might be keeping the $i$th eigenvector $\nu$ of A as the smallest eigenvector of B, and invent all $N1$ eigenvectors of B. It is equivalent to find an arbitrary set of orthogonal vectors in the $N1$ 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 22nd, 2015 at 12:51 AM. 
September 25th, 2015, 06:41 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
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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definite, eigenvalue, eigenvector, matrix, positive, predefined, smallest 
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