Let $M$ be a real $3 \times 3$ symmetrix matrix with positive entries with one eigenvalue equal to 1 and all others equal to 0. In other words, the matrix can take the form $M=v v^\top$ for some real $v^\top = (\sqrt{v_1},\ldots \, \sqrt{v_3})$ such that $\sum_i v_i =1$. Assume that I make a small perturbation of each element so that $(1-\delta^2)\leq \frac{M_{ij}}{\sqrt{v_i v_j}}\leq 1$, where $\delta \ll 1$ is some sufficiently small number. What can I say about the maximum eigenvalue of the new matrix and can I lower or upper bound it?
For example, if all the elements of the new matrix $M^\prime_{ij}=(1-\delta^2){\sqrt{v_i v_j}}$ that would imply the maximum eigenvalue is $1-\delta^2$ and all the rest 0, which however is not possible in the exact problem I have since the sum of the eigenvalues should be 1. That is, the diagonal elements do not change and the trace remains 1.
In all generality, the perturbation I am considering is of the form $$M^\prime_{ij}= M_{ij} \left( \sqrt{\left(\frac{1}{2}+\delta_i\right)\left(\frac{1}{2}+\delta_j\right)} +\sqrt{\left(\frac{1}{2}-\delta_i\right)\left(\frac{1}{2}-\delta_j\right)} \right),$$ for some $0\leq\delta_i\leq \delta$ which are in general non equal for different $i$. A simple expansion of the function in brackets up to quadratic terms would imply that $$ M^\prime_{ij} \sim M_{ij} \left(1-\left(\delta_i-\delta_j\right)^2\right)\geq M_{ij}(1- \delta^2) $$ I would like to carefully provide a bound on the largest eigenvalue, $e_0(M^\prime)$, of $M^\prime$ and it seems that it should be something like $e_0(M^\prime) \leq 1-\delta^2$, but I can't show it properly.
Any ideas or known results on the topic? Thank you!