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Denote the number of positive eigenvalues of a Hermitian matrix $H$ by $P_H$. If $A,B$ Hermitian, show that $$P_{A+B}\leq P_A+P_B.$$

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    $\begingroup$ For future googlers: This property is known as the Subadditivity of Inertia. Also, it's also true if you let P denote the number of negative eigenvalues. (This is easy to prove by substituting -A for A and -B for B) (And it's NOT true if you let P denote the number of eigenvalues equal to zero.) $\endgroup$ – ahuff44 Mar 22 '14 at 8:45
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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 $j=P_B$.

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