TITLE: Matrix trace & norm
QUESTION [14 upvotes]: 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?

REPLY [9 votes]: 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 
$$
tr(AB) \leq|tr(AB)|=\left|\sum a_{ii}b_{ii}\right| \leq \sum a_{ii}|b_{ii}| \leq \sum a_{ii} \|B\|=tr(A)\|B\|.
$$