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Here is a first-order approximation. I'll write $\tilde{v} = v + \rho w + O(\rho^2)$ and $\tilde{\lambda} = \lambda + \rho \mu + O(\rho^2)$. We may assume $\|v\| = \|\tilde{v}\| = 1$, so $v^T w = 0$. If $\lambda$ is a simple eigenvalue of $B$, $B - \lambda I$ is invertible on the orthogonal complement $v^\perp$ of $v$. Taking $(B + \rho b b^T) \tilde{v} = \tilde{\lambda} \tilde{v}$ to first order in $\rho$, we have $$ B w + b b^T v = \mu v + \lambda w $$ Let $c = b - (b^T v) v$ which is orthogonal to $v$, so that this equation splits into $$ B w + (b^T v) c = \lambda w \ \text{and} (b^T v)^2 v = \mu v$$ Thus $\mu = (b^T v)^2$ and $w = - (b^T v) (B - \lambda I)|_{v^\perp}^{-1} c$.