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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

NaSa
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Jul 28 '21 at 19:55
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    $B^{1/2}AB^{1/2}$ is positive semidefinite and nonzero. Hence its trace is positive. Consequently, $\operatorname{tr}(AB)=\operatorname{tr}(B^{1/2}AB^{1/2})>0$. – user1551 Jul 29 '21 at 07:57
  • Hi user1551. Thank you for the quick reply! I see why the trace of $B^{1/2}AB^{1/2}$ is nonnegative (as an implication of semidefinitenes). However, could you tell please me why the fact that $B^{1/2}AB^{1/2}$ is nonzero implies that the trace of $B^{1/2}AB^{1/2}$ is always positive? – NaSa Jul 29 '21 at 08:05
  • Hi @daw. Thanks for the link. Unfortunately, that does not help. It only solves part of the problem, however it remains to show that $tr(AB)\neq 0$, which I dont know. In any case thanks for the suggestion – NaSa Jul 29 '21 at 12:12
  • @NaSa The trace of a positive semidefinite matrix is the sum of its eigenvalues. If the matrix is nonzero then there is at least one strictly positive eigenvalue and hence the trace is strictly positive. – Rammus Jul 29 '21 at 16:14
  • @Rammus Unfortunately this is not the case. Here is the example non zero matrix $\left( \begin{matrix}0 & 1\ 0 & 0\end{matrix}\right)$ has zero eigenvalues. – NaSa Jul 30 '21 at 09:04

1 Answers1

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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 $\text{rank}\big(A^\frac{1}{2}B^\frac{1}{2}\big) =\text{rank}\big(A^\frac{1}{2}\big)\neq 0$

user8675309
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