As the title says, for $A,B\in \mathbb{S}^n$, $B\ge 0$ (PSD), how to prove that $$ \lambda_{\min} (A)\operatorname{tr}(B)\le \operatorname{tr}(AB) \le \lambda_{\max} (A)\operatorname{tr}(B). $$
Is it possible to use eigendecomposition to prove this?
For any matrix M, and basis $v_i$, $tr(M) = \Sigma v_i^T M v_i$. Apply this to a basis of eigenvectors for A (which exists because A is real and symmetric) to compute $tr(AB)$. You will have to use B PSD, in particular $v^T B v$ nonnegative, to get the inequalities to go the right direction.
That is, let $v_i$ be a basis for $A$ with the property that $A v_i = \lambda_i v_i$. Note that because $A$ is symmetric, we also have $v_i^T A = \lambda v_i^T$
Then we write:
$tr(AB) = \Sigma_{i = 1}^n v_i^T AB v = \Sigma_{i = 1}^n \lambda_i v_i^T B v_i \leq \Sigma \lambda_{max} v_i^T B v_i = \lambda_{max} \Sigma v_i^T B v_i = \lambda_{max} tr(B)$.
(Note that for the middle inequality, we needed that $v_i^T B v_i \geq 0$.)
The case for $\lambda_{min}$ is similar.
I assume the assertion $A, B \in \Bbb S^n$ is to be interpreted as meaning that both $A$ and $B$ are real, symmetric, $n \times n$ matrices. I further assume the assertion that $B \ge 0$, i.e. $B$ is a PSD (positive semidefinite) matrix means that for any real vector $x$, $x^TBx \ge 0$.
Since $A \in \Bbb S^n$, there is an orthogonal matrix $O$, that is,
$OO^T = O^TO = I, \tag 1$
which diagonalizes $A$; we can then set
$O^TAO = \Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n) = [\lambda_i \delta_{ij}]; \tag 2$
here the $\lambda_i \in \Bbb R$ are of course the eigenvalues of the matrix $A$.
We recall that matrix traces are invariant under similarity transformations such as $A \to O^TAO = O^{-1}AO$; therefore
$\operatorname{tr}(AB) = \text{tr}(O^TABO) = \text{tr} (O^TAOO^TBO), \tag 3$
where we have used (1) on the right of (3); if we write the transformed matrix $O^TBO$ as
$O^TBO = [b_{ij}], \tag 4$
we may combine (2), (3) and (4) and obtain
$\text{tr}(AB) = \text{tr}(O^TAOO^TBO) = \text{tr}(\Lambda [b_{ij}]) = \text{tr}([\lambda_i \delta_{ik}][b_{kj}]) = \text{tr}([\lambda_i b_{ij}]) = \displaystyle \sum_1^n \lambda_i b_{ii}. \tag 5$
Now, since the columns $\vec e_i$ of an orthogonal matrix such as $O$ form an orthonormal set,
$O = [\vec e_1 \; \vec e_2 \; \ldots \: \vec e_n ], \tag 6$
a fact which is manifested in (1), where
$O^T = \begin{bmatrix} \vec e_1^T \\ \vec e_2^T \\ \vdots \\ \vec e_n^T \end{bmatrix}, \tag 7$
we have
$[b_{ij}] = O^TBO = [\vec e_i^TB\vec e_j], \tag 8$
from which it is seen that
$b_{ii} = \vec e_i^T B \vec e_i \ge 0 \tag 9$
since $B$ is positive semidefinite; it follows then that
$\displaystyle \lambda_{min} \sum_1^n b_{ii} = \sum_1^n \lambda_{min} b_{ii} \le \sum_1^n \lambda_i b_{ii} \le \sum_1^n \lambda_{max} b_{ii} = \lambda_{max} \sum_1^n b_{ii}, \tag{10}$
whence
$\lambda_{min} \text{tr}(B) \le \text{tr}(AB) \le \lambda_{max} \text{tr}(B); \tag{11}$
HINT:
Use the following: if $C$ and $B$ are $PSD$ then $\operatorname{tr}(CB)\ge 0$. Indeed, $\operatorname{tr}(CB)=\operatorname{tr}(\sqrt{B}C\sqrt{B})\ge0$..
Now, $\lambda_{\max}(A)\cdot I-A \succeq 0$ so $$\operatorname{tr}(\, (\lambda_{\max}(A)\cdot I-A)B\,) \ge 0$$ Expand and get the RHS inequality .
Yes. Eigendecomposing $A$ we obtain $A = U D U^T$, where $U$ is an orthogonal matrix and $D$ a diagonal matrix composed by the eigenvalues of $A$. Now note that $$ \operatorname{tr}(A) = \operatorname{tr}(U D U^T) = \operatorname{tr}(D) $$ $$ \operatorname{tr}(B) = \operatorname{tr}(U^T B U) $$ $$ \operatorname{tr}(AB) = \operatorname{tr}(U^T A B U) = \operatorname{tr}(D U^T B U) $$ and $U^T B U \ge 0 $, since $B \ge 0$. Then all the diagonal elements of $U^T B U$ are nonnegative. Thus $$ \lambda_{\min}(A)\operatorname{tr}(B) = \lambda_{\min}(D)\operatorname{tr}(U^T B U) \le \operatorname{tr}(D U^T B U) = \operatorname{tr}(AB) $$ and $$ \operatorname{tr}(AB) = \operatorname{tr}(D U^T B U) \le \lambda_{\max}(D)\operatorname{tr}(U^T B U) = \lambda_{\max} (A)\operatorname{tr}(B) $$
