# Inequalities about the eigenvalues of real symmetric matrices $A,B,A+B$

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 $$AB\ne BA$$?

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)