inequality of trace of matrices

Question

Suppose that the $n \times n$ matrices $A$ and $B$ are both symmetric positive definite and satisfy $A > B$ (meaning that $A - B$ is positive definite).

1. If a symmetric positive definite matrix $C$ is multiplied from the right, then is the following true: $AC > BC$?
2. Furthermore, does the following result hold: $\text{Trace}(AC) > \text{Trace}(BC)$?

Answer 1

1) Let $D = A - B$. So given $D > 0$ and $C > 0 $, you are seeking if $DC > 0$ is true. The product of two positive definite matrices is not necessarily positive definite (see here).

2) $\operatorname{trace}(AC)>\operatorname{trace}(BC)$ is same as asking $\operatorname{trace}(DC)>0$. Now let $C = R R^T$ (Cholesky decomposition). Then, $$ \operatorname{trace}(DC) =\operatorname{trace}(R^TDR) =\sum_{i} r_i^T D r_i, $$ where $r_i$ are columns of $R$.