Let $A$ be a real symmetric matrix whose spectrum is $\lambda_1,\lambda_2,\ldots,\lambda_n$. Let $A'$ be the matrix obtained by adding a perturbation to $A$. The requirement is that only the second eigenvector of $A'$ is known and fixed in advance. Is there a way to obtain the set of all possible perturbations which will guarantee that $A'$ has the specified second eigenvector?

1$\begingroup$ It is not too hard to characterize rank one perturbations which make a given vector an eigenvector. I assume what makes the problem difficult is to guarantee that this eigenvector is the "second" one. You did not state the criterium for ordering, however. $\endgroup$ – Michael Renardy May 31 '11 at 15:03

$\begingroup$ To clarify, can you confirm: $A'=A+\epsilon H$; $A$ and $H$ are symmetric and $H$ is rankone, thus $H=u\otimes u$. The second eigenvector of $A'$ is prescribed. It is not necessarily the second eigenvector of $A$, though. Also, it seems implicitly assumed that the second eigenvalue (in decreasing order?) of $A'$ is simple. $\endgroup$ – Pietro Majer May 31 '11 at 18:22
The perturbation matrix $\Pi$ must have $v_2$ (your second eigenvector) as an eigenvector. The associated eigenvalue is either $0,$ or not. In the second case, $\Pi = c v_2 v_2^t$ (in other words, a multiple of the projection onto the space spanned by $v_2.$ In the first case, $\Pi = c w w^t,$ where $(w, v_2) = 0$ (in English, a multiple of a projection onto some onedimensional subspace orthogonal to $v_2.$ As for the ordering of the eigenvalues (as mentioned by @Denis), this is very easy to check in the first case ($c < \lambda_1  \lambda_2$), and slightly less easy in the second case (you need to compute the projection of $w$ onto $v_1$).
@Pietro Majer,
A few clarifications in the question. I am interested in the case of the matrix being the Laplacian of some undirected weighted graph. Thus, A would now be real,symmetric and positive semi definite. Also A' = A + \epsilon H where H is rankone. Importantly the second eigenvector i meant was the one the corresponds to the second "smallest" eigenvalue.
Thanks.