Lower bounds on eigenvalues of sum of two matrices (one of them is symmetric)

Question

Let $L$ be a matrix with eigenvalues($\lambda$ $\geq$ 0). If I add a constant value (say $a$) to all the elements of $L$, what can we say about the minimum eigenvalue of this perturbed matrix? Note: $L$ has precisely one zero eigenvalues.

Answer 1

$\newcommand\la\lambda\newcommand\1{\mathbf1}$There is no finite lower bound in terms of $a$ and the eigenvalues of $L$ on the minimum eigenvalue of the perturbed matrix $L+a\,\1\1^\top$ given only the information you gave; here, $\1\1^\top$ is the $2\times2$ matrix of $1$'s.

Indeed, let $$L=\begin{bmatrix}\la&b\\0&0\end{bmatrix}$$ for some real $\la\ge0$ and $b$. Then the eigenvalues of the perturbed matrix $L+a\,\1\1^\top=\begin{bmatrix}\la+a&b+a\\a&a\end{bmatrix}$ are $$t_\pm:=\la/2+a\pm\sqrt{(\la/2+a)^2+(b-\la)a},$$ so that for any given real $\la\ge0$ and $a>0$ we have $t_-\to-\infty$ as $b\to\infty$.

Also, for any given real $\la\ge0$ and $a>0$, if $b$ is close enough to $-\infty$, then the eigenvalues $t_\pm$ are not real, and hence it makes no sense to even talk about the minimum eigenvalue of the perturbed matrix.