# Positive definite matrix whose smallest eigenvector (with the smallest eigenvalue) 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.

• I am not quite sure what you are asking. Do you just want to construct a positive definite matrix with prescribed eigenvalues? – Tobias Kildetoft Sep 22 '15 at 7:46
• I want to construct a positive definite matrix with a prescribed eigenvector, and this eigenvector must have the smallest eigenvalue. – whitegreen Sep 22 '15 at 7:49
• @whitegreen Is it possible to take the vector, normalize it, complete to an ON basis, that is to a unitary matrix $P$, pick any positive diagonal $\Omega$ with $\Omega_1$ being the smallest and set $A=P\Omega P^*$? – A.Γ. Sep 22 '15 at 7:57
• @A.G. Yes, it works! Modifying the eigenvalues in $\Omega$ to let the prescribed eigenvector's eigenvalue be the smallest, then set $A=P \Omega P^{-1}$, done!. – whitegreen Sep 23 '15 at 3:07