Prove that $\lambda_{\max}(A) +\lambda_{\min}(B) \leq \lambda_{\max}(A+B) \leq \lambda_{\max}(A) + \lambda_{\max}(B).$ Let $A,B \in \mathbb{M}_{n\times n}(\mathbb{C})$ be Hermitian matrices. Prove that $$\lambda_{\max}(A) +\lambda_{\min}(B) \leq \lambda_{\max}(A+B) \leq \lambda_{\max}(A) + \lambda_{\max}(B).$$
Solution given: Let $\lambda = \lambda_{\max}(A+B)$ and v an associated eigenvector such that $||v|| =1$. Then (here)$\lambda = v^*(\lambda v) \leq v^*(A+B)v = v^*Av + v^*Bv \leq \lambda_{\max} A + \lambda_{\max} B$. Similarly, $\lambda_{\max}A \leq \lambda_{\max}(A+B) + \lambda_{\max}(-B) = \lambda_{\max}(A+B) - \lambda_{\min}B.$ So $\lambda_{\max}(A+B) \geq \lambda_{\max} A + \lambda_{\min} B$.
My question is why $\lambda_{\max}(A+B) = v^*(\lambda v)$? If it is not, why the prove is proving that $v^*(\lambda v)$ is in the range? The prove has previously prove that it is true for $v^*(\lambda v)$.
 A: Remember that $\|v\|^2 = \langle v,v \rangle = v^*v = 1$. 
It follows that $v^*(\lambda v) = \lambda (v^*v)= \lambda$.
A: I'll use inner product notation $(u,v)=v^{\star}u$.
If $A$ is Hermitian, then function $f(v)=(Av,v)$ is a real continuous function on the unit sphere $S=\{ v \in \mathbb{C}^{n} : \|v\|=1\}$. Its minimum value is $f_{\min}=\lambda_{\min}(A)$, which it achieves iff $v$ is a unit vector with $Av=\lambda_{\min}(A)v$. Similarly, the maximum values of $f$ is $f_{\max}=\lambda_{\max}(A)$, which is achieved at any unit vector $v$ for which $Av=\lambda_{\max}(A)v$. This is true of any Hermitian matrix $A$. To see why this is true, recall that $A$ has an orthonormal basis of eigenvectors $\{ e_{k} \}_{k=1}^{n}$, which can be numbered so that the corresponding eigenvalues are
$$
      \lambda_{\min}=\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n \le \lambda_{\max}.
$$
Some eigenvalues may be duplicated in this list; so the ordering is not necessarily unique. Every unit $v$ can be expanded as $v=\sum_{k}(v,e_{k})e_{k}$ with $\|v\|^{2}=\sum_{k}|(v,e_{k})|^{2}$. Furthermore,
$$
                         (Av,v) = \sum_{k}\lambda_{k}|(v,e_{k})|^{2}\\
            \lambda_{\min}\sum_{k}|(v,e_{k})|^{2} \le (Av,v) \le \lambda_{\max}\sum_{k}|(v,e_{k})|^{2}
$$
If $v$ is a unit vector, then $\lambda_{\min} \le (Av,v)\le \lambda_{\max}$. The minimum of $f(v)=(Av,v)$ is achieved at the unit vecotr $v=e_{1}$ and the maximum is achieved at $v=e_{n}$.              
If $A$ and $B$ are Hermitian, then $A\pm B$ are Hermitian as well. So the above discussion applies equally well to all of these functions
$$
     f(v)=(Av,v),\;\; g(v)=(Bv,v),\\ (f+g)=((A+B)v,v),\;\; (f-g)(v)=((A-B)v,v).
$$
But the thing to be remembered is that a unit vector which minimizes $(f+g)$ may not be a point which minimizes either $f$ or $g$. There is a point $v \in S$ where $f+g$ is minimized; it is minimized at any unit eigenvector $w$ with eigenvalue $\lambda_{\min}(A+B)$. There's no reason to believe, however, that $w$ is an eigenvector of $A$ or of $B$, which is to say that there is no reason to believe that $f$ has its minimum at $w$ or that $g$ has its minimum at $w$ just because $f+g$ has its minimum at $w$.
