Lower bound on the minimum eigenvalue of sum of two matrices

Question

Assume that $A$ is a symmetric positive definite matrix and $B$ is a symmetric (can potentially have negative entries). Is the following bound correct?

$$ \lambda_{\min}(A+B) \geq \lambda_{\min}(A) + \lambda_{\min}(B) $$

Answer 1

Assuming that $A,B$ are Hermitian ($A^* = A, B^*=B$) then the eigenvalues are real and satisfy $\langle x, Ax \rangle \ge \lambda\|x\|^2$ for all $x$ iff $\lambda_\min(A) \ge \lambda$.

To see this, note that $\langle x, Ax \rangle \ge \lambda_\min(A)\|x\|^2$ for all $x$, and that the bound is attained for an eigenvector corresponding to the eigenvalue $\lambda_\min(A)$.

We have $\langle x, (A+B)x \rangle = \langle x, Ax \rangle + \langle x, Bx \rangle \ge \lambda_\min(A) \|x\|^2 + \lambda_\min(B) \|x\|^2$ for all $x$, from which the desired result follows.

Answer 2

The bound also follows from an application of Weyl's inequalities (sometimes called the Courant-Weyl inequalities).