Matrix trace & norm For any nonnegative semidefinite matrix $A$ and any matrix $B$, we have
$$\mbox{tr} (AB) \le \mbox{tr} (A) \, \|B\|$$
where $\mbox{tr}(\cdot)$ is the trace and $\|\cdot\|$ is the operator norm. How to prove this?
 A: We can work in an orthonormal basis in which $A$ is diagonal. In that case the diagonal entries of $AB$ are $a_{ii}b_{ii}$, so 
$$
tr(AB) \leq|tr(AB)|=\left|\sum a_{ii}b_{ii}\right| \leq \sum a_{ii}|b_{ii}| \leq \sum a_{ii} \|B\|=tr(A)\|B\|.
$$
A: A more general version of the result Beenakker cites that allows for normal $B$ (not just symmetric):$\newcommand\tr{\operatorname{tr}}\newcommand\E{\mathbb{E}}$
Take $x\sim\mathcal{N}(0,I)$ and recall that $\tr(C) = \E[x^\mathsf{T}Cx]$. If $B = Q \Lambda Q^*$ is unitarily diagonalizable let $\Lambda^{1/2}$ be a (possibly complex) square root of $\Lambda$. Then using the cyclic property of the trace
$$
\begin{align}
\tr(AB) &= \tr(\Lambda^{1/2} Q^*AQ\Lambda^{1/2})\\
&= \E[(\Lambda^{1/2}x)^\mathsf{T} Q^*AQ (\Lambda^{1/2}x)]\\
&= \sum_{i\neq j} \lambda_i^{1/2}\lambda_j^{1/2}\E[x_ix_j](Q^*AQ)_{ij} + \sum_{i} \lambda_i\E[x_i^2](Q^*AQ)_{ii}\\
&= \sum_{i} \lambda_i(Q^*AQ)_{ii}.
\end{align}
$$
From here, since $(Q^*AQ)_{ii}\geq 0$ by the PSD condition, it follows that
$$\tr(AB) \leq \lambda_{\max}\sum_{i} (P^{-1}A P)_{ii} = \lambda_{\max}\tr(P^{-1}AP) = \lambda_{\max}\tr(A)$$
and
$$\tr(AB) \geq \lambda_{\min}\sum_{i} (P^{-1}A P)_{ii} = \lambda_{\min}\tr(P^{-1}AP) = \lambda_{\min}\tr(A)$$
as desired.
A: For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative trace. This gives $\mathrm{tr}(A(\|B\| - B)) \geq 0$, from which the inequality follows immediately.
For general $B$, we can therefore argue that
$$\mathrm{Re}(\mathrm{tr}(AB)) = \tfrac{1}{2}\mathrm{tr}(A(B + B^*))\leq \mathrm{tr}(A)\cdot \tfrac{1}{2}\|B + B^*\| \leq \mathrm{tr}(A) \|B\|.$$
Upon multiplying $B$ by a suitable complex scalar, this implies that
$$|\mathrm{tr}(AB)| \leq \mathrm{tr}(A) \|B\|$$
is true for all positive semidefinite $A$ and any $B$.
I don't have the book at hand right now, but I imagine that this is somewhere to be found in the early chapters of Bhatia's Matrix Analysis.
A: Expanding my comment into an answer, which offers a more general result.

Theorem (von Neumann). Let $A$ and $B$ be arbitrary $n\times n$ complex matrices. Then, $$|\text{trace}(AB)| \le \sum_{i=1}^n \sigma_i(A)\sigma_i(B),$$ where $\sigma_i(\cdot)$ denotes the $i$-th singular value.

As an immediate corollary we have the OP's result, because for a PSD matrix $A$, $\sigma_i(A)=\lambda_i(A)$.

Corollary. Let $A$ be Hermitian positive semidefinite, and $B$ be arbitrary. Then, $|\text{trace}(AB)| \le \text{trace}(A)\|B\|$, where $\|\cdot\|=\sigma_1(\cdot)$ denotes the operator norm.


Note. Using von Neumann's trace inequality, we can also show the more general version of the OP's question, namely a Hölder inequality:
\begin{equation*}
|\text{trace}(AB)| \le \langle \sigma(A), \sigma(B)\rangle \le \|\sigma(A)\|_p\|\sigma(B)\|_q = \|A\|_p\|B\|_q,
\end{equation*}
where $\sigma(A)$ is the vector of singular values, and $\|\cdot\|_p$ denotes the Schatten $p$-norm, and $1/p + 1/q=1$ ($p,q\ge1$). For $p=1$ and $q=\infty$ we recover the OP's question.
A: A proof of
$$\lambda_n(B)\,{\rm tr}\,A\leq {\rm tr}\,(AB)\leq\lambda_1(B)\,{\rm tr}\,A$$
where $\lambda_n$ is the $n$-th largest eigenvalue of $B$ so $||B||=\lambda_1(B)$ and $A$, $B$ are positive semidefinite $n\times n$ matrices, is given in The design of suboptimal linear time-varying systems (1968). The inequality still holds for $B$ real symmetric and $A$ positive semidefinite, as proven in Trace bounds on the solution of the algebraic matrix Riccati and Lyapunov equation (1986). The restriction of positive semidefinite $A$ cannot be relaxed, as pointed out in Inequalities for the trace of a matrix product (1994).
