Let $A, B$ be symmetric matrics. Define
$$
N(A):=\sum_{\lambda (A)<0}\lambda (A) ,\qquad P(A):=\sum_{\lambda (A)>0} \lambda (A) $$,
where $A$ is a real symmetric matrix, $\lambda(A)$ represents its eigenvalue and eigenvalues whose algebric multiplicity is $k$ are sumed $k$ times.
Question: How to prove the following inquality?
$$
N(A)+N(B)\le N(A+B)\le P(A)+N(B)\le P(A+B)\le P(A)+P(B)
$$
My effort:When $A,B$ are diagonal matrics, we can use the following equality easily get a proof.
$$
\begin{pmatrix}
\lambda_1& & & & \\
& \lambda_2& &\\
& & \ddots& &\\
& & & \lambda_n&\\
\end{pmatrix}+\begin{pmatrix}
\mu_1& & & & \\
& \mu_2& &\\
& & \ddots& &\\
& & & \mu_n&\\
\end{pmatrix}=\begin{pmatrix}
\lambda_1+\mu_1& & & & \\
& \lambda_2+\mu_2& &\\
& & \ddots& &\\
& & & \lambda_n+\mu_n&\\
\end{pmatrix}
$$
where $\lambda_i,\mu_i$ respectively deontes $A,B$'s eigenvalues. Hence, when $AB=BA$, above inequlities also true. Because $A,B$ can be expressed by $P^{T} \Lambda_1 P,P^{T}\Lambda_2 P$, where $\Lambda_1,\Lambda_2$ are digonal matrics and $P$ is a orthogonal matrix, namely $PP^{T}=I$.
Question: How to prove these inequlities when $AB\ne BA$?
Best Answer
For background reference: Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$ .
lemma
with real symmetric $X, Y$
$P\Big(\big(X+Y\big)\Big)\leq P\big(X\big)+P\big(Y\big)$
where $(X+Y)$ has $r$ eigenvalues $\gt 0$ and for some real symmetric idempotent $S$ with rank $r$
$P\Big(\big(X+Y\big)\Big)$
$= \text{trace}\Big(S\big(X+Y\big)\Big)$
$= \text{trace}\Big(SX\Big) +\text{trace}\Big(SY\Big)$
$\leq \big(\sum_{k=1}^r \lambda_k(X)\big)+\big(\sum_{k=1}^r \lambda_k(Y)\big)$
$\leq P\big(X\big)+P\big(Y\big)$
main inequalities
1.) $P(A+B)\leq P(A)+P(B)$ by $X:=A$ and $Y:=B$
2.) The final, left most one is better written by negating and seeing it as $P(-A+-B) \leq P(-A) + P(-B)$ i.e. it is the same as the (1) just using negation
3.) $P(A)+N(B)\le P(A+B)$ is better written as $P(A)\leq P(A+B) + P(-B)$ with $X:=A+B$ and $Y:=-B$. Note $Y:=-A$ gives $P(B)\leq P(A+B) + P(-A)$
4.) finally: $N(A+B)\leq P(A)+N(B)$, but this is just (3) after exploiting negation.
i.e. if we negate it we get the equivalent claim $-P(A)-N(B)= N(-A)+P(-B)\leq P\big((-A+-B)\Big)=-N(A+B) $ which holds by (3)