Proof that trace is non-negative: positive definite matrix and a vector

Question

Let $A$ be a positive definite matrix and $b$ be some vector. Then show \begin{align} \textrm{Tr}(A^{-1}bb^T A{-1})\geq 0 \end{align} I'm not sure how to show this. Any tips? It's from p. 13 of https://arxiv.org/pdf/1602.02845.pdf.

Answer 1

WTS: $tr(A^{-1}bb^TA^{-1})$ (as stated in the paper)

First $A$ positive definite implies $A^{-1}$ positive definite. Define $y = b^TA^{-1}$. Then using the commonly assumed fact that positive definite implies symmetric, we have

$$ tr(A^{-1}bb^TA^{-1}) = tr(y^Ty) \geq 0 $$

where the inequality follows because $y^Ty = \sum_{i=1}^n y_i^2$.