Suppose diagonal $D\in \mathbb R^{n\times n}$ with $D\succeq 0, v\in \mathbb R^n,$ and $\alpha>0$ are given. Can we $\textit{exactly}$ identify the smallest eigenvalue of $D+\alpha vv^T$ and its corresponding eigenvector (based on $v$ and given eigenpairs of $D$)?
Even though the method to identify the whole spectrum is given in Eigenvectors of rank one update matrix, I am only interested in finding the smallest eigenvalue and its eigenvector. So, I wonder if something more can be said!
Best Answer
I think the only simple connection that can be shown for the smallest eigenvalue of $D+\alpha vv^T$ is the following:
$$\lambda_{\min}(D+\alpha vv^T) \ge \min (d_1,\dots,d_n)=\lambda_{\min}(D).$$
This follows from the fact that the roots of the following
$$f(\lambda) = 1+\alpha \sum_{i=1}^n \frac{v_i^2}{d_i-\lambda}$$
are the eigenvalues of $D+\alpha vv^T$ (specials cases of $v_i=0$ or $d_i=d_j$ for some $i,j$ can be managed following this answer). One can see that $f(\lambda)$ tends to $1$ and $+\infty$ as $\lambda \to -\infty$ and $\lambda \to \left (\min ( d_1,\dots,d_n ) \right )^-$, respectively. As $f(\lambda)$ is increasing over this interval, the smallest root of $f(\lambda)$ cannot be smaller than $\min ( d_1,\dots,d_n )$; it can be equal to if $v_i=0$ for $i$ with $d_i=\min ( d_1,\dots,d_n )$.
This could be alternatively proven using the fact that $\lambda_{\min}(X)$ is increasing in $X \in S^n$ with respect to the order $\succeq $ for any symmetric $D$.
In a very special case that $d_1\ne 0, d_2=\dots=d_n=0$, the eigenvalue can be obtained explicitly (which is also known from here for more general form of $uu^T+vv^T$).
After finding the eigenvalue $\tilde \lambda=\lambda_{\min}(D+\alpha vv^T)$, the eigenvector $\tilde q$ can be explicitly obtained using the Bunch–Nielsen–Sorensen formula developed for this purpose as follows:
$$\tilde q_i = c\frac{v_i}{d_i - \tilde \lambda}, i=1,\dots,n $$
where $c\neq 0$ is an arbitrary number.