In the paper by Bai and Ng (2002) I came across the following claim:
Problem: Consider $(n\times n)$ matrix $A$, which is symmetric positive semi-definite and $A\neq 0$, and matrix $B$ is $(n\times n)$ positive definite. This implies $tr(AB)>0$?
Comment 1: Unfortunately I failed to see why this is the case. My first thought to solve this problem was to use the following result: If $B$ is positive definite $(n \times n)$ matrix and $(n \times m)$ matrix $A$ has $rank(A)=m$ then $A'BA$ is positive definite. (see e.g., Lutkepohl, 1996, Handbook of Matrices, fact 9.12.1 (10)). However, the rank of $A$ in the original problem cannot be evaluated. Any suggestions why/when this claim is true are highly appreciated
Comment 2: It is follows immediately that $tr(AB)\geq 0$. However, it is not clear to me how one can show that $tr(AB)\neq 0$. (Note that nonzero matrix can have all zero eigenvalues)
References
Problem that I mention is in the appendix of the paper Lemma 3, top of the page 217.
Bai, J. and S. Ng (2002). Determining the number of factors in approximate factor models. Econometrica 70(1), 191–221
working over reals, we can assume WLOG that $B$ is symmetric PD
if not, because transposition doesn't change trace or positive definiteness, $\text{trace}\Big(AB\Big) = \text{trace}\Big(A\big(\frac{1}{2}(B+B^T)\Big)$
and working with $C:=\frac{1}{2}(B+B^T)\succ \mathbf 0$ gives the result
main argument
symmetric PSD matrices have a square root
(if you prefer, you can use Cholesky or $LDL^T$ factorization instead, it doesn't matter)
$\text{trace}\Big(AB\Big) =\text{trace}\Big(A^\frac{1}{2}A^\frac{1}{2}B^\frac{1}{2}B^\frac{1}{2}\Big)=\text{trace}\Big(B^\frac{1}{2}A^\frac{1}{2}A^\frac{1}{2}B^\frac{1}{2}\Big)=\big\Vert A^\frac{1}{2}B^\frac{1}{2}\big\Vert_F^2\gt 0 $
by positive definiteness of the (squared) Frobenius norm and because $\text{rank}\big(A^\frac{1}{2}B^\frac{1}{2}\big) =\text{rank}\big(A^\frac{1}{2}\big)\neq 0$