TITLE: Subadditivity of positive index of inertia
QUESTION [8 upvotes]: Denote the number of positive eigenvalues of a Hermitian matrix $H$ by $P_H$. If $A,B$ Hermitian, show that $$P_{A+B}\leq P_A+P_B.$$

REPLY [7 votes]: This is a corollary of the Weyl's theorem on eigenvalues. One of its statements says that
$$
\lambda_i(A+B)\leq\lambda_{i+j}(A)+\lambda_{n-j}(B)
\qquad\text{for $j=0,\ldots,n-i$},
$$
where $A,B$ are $n\times n$ real symmetric or complex Hermitian and the eigenvalues are ordered so that $\lambda_1(\cdot)\leq\cdots\leq\lambda_n(\cdot)$. You want to show that $P_{A+B}\leq P_A+P_B$, which is the same as
$$\lambda_{n-(P_A+P_B)}(A+B)\leq 0.$$
To show that use the above inequality with $i=n-(P_A+P_B)$ and $j=P_B$.