Denote the number of positive eigenvalues of a Hermitian matrix by $P(\cdot)$. If $A,B$ Hermitian, show that $$P(A+B)\leq P(A)+P(B).$$
I know there is an elementary proof:
Subadditivity of positive index of inertia
But my teacher says that this problem has other elementary proofs. Can anybody help? Thank you.
As far as I know, this is the most elementary proof based on the spectral theorem for Hermitian matrices and a simple dimensional argument.
Let $V_+(X)\subset\mathbb{C}^n$ be the subspace spanned by the eigenvectors corresponding to the positive eigenvalues of a Hermitian matrix $X$ and $V_+(X)^{\perp}$ its orthogonal complement (hence the subspace spanned by the eigenvectors corresponding to the non-positive eigenvalues). We have $$\tag{$*$}V_+(A)+V_+(B)+V_+(A+B)^{\perp}=\mathbb{C}^n.$$ Hence the sum of the dimensions of these three spaces cannot be smaller than $n$: $$ n\leq\dim V_+(A)+\dim V_+(B)+\dim V_+(A+B)^{\perp}. $$ Since $\dim V_+(A)=P(A)$, $\dim V_+(B)=P(B)$, and $\dim V_+(A+B)^{\perp}=n-P(A+B)$, we have $$ n\leq P(A)+P(B)+(n-P(A+B)) \quad \Leftrightarrow \quad P(A+B)\leq P(A)+P(B). $$
To see why ($*$) holds, note that it is equivalent (take the complement) to $$V_+(A)^{\perp}\cap V_+(A)^{\perp}\cap V_+(A+B)=\{0\}.$$ If this was not true, there would be a nonzero $x$ such that $x^*Ax\leq 0$ and $x^*Bx\leq 0$, and $x^*(A+B)x>0$. This is, however, not possible.
For the reference, see this paper.