How to show ${\rm tr}[AXA'] \geq {\rm tr}[X]$ for $X\geq 0$ if $AA' \geq I$?

Question

Assume that $A \in \mathbb{R}^{n \times n}$ and $AA' \geq I_n$, where $A'$ is the transpose of $A$, and $X \in \mathbb{R}^{n \times n}$ and $X \geq 0$, i.e., $X$ is a positive matrix . How to show \begin{equation} {\rm tr}[AXA'] \geq {\rm tr}[X], \end{equation} where ${\rm tr}[X]$ is the trace of $X$, i.e., ${\rm tr}[X] = \sum_{i=1}^n X_{ii}$.

Answer 1

Define matrix $ M = \begin{bmatrix} I_n & -A \\ A' & I_n \end{bmatrix} $, since $ I_n $ is invertible, thus we have \begin{equation} \det[M] = \det[I_n] \det[I_n - A'I_n^{-1}(-A)] = \det[I_n +A'A]. \end{equation} On the other hands, we have \begin{equation} \det[M] = \det[I_n] \det[I_n - (-A)I_n^{-1}A'] = \det[I_n +AA'], \end{equation} thus there must be that $ \det[I_n +A'A] = \det[I_n +AA']$. Let $ \lambda \neq 0 $, then we have \begin{equation}\label{key} \det[AA' - \lambda I_n] = (-\lambda)^n \det[\frac{AA'}{\lambda}+I_n] = (-\lambda)^n \det[\frac{A'A}{\lambda}+I_n] = \det[A'A - \lambda I_n], \end{equation} this means that $ AA' $ and $ A'A $ have the same nonzero eigenvalues.

Note that $ AA' \geq I_n $, thus we have $ A'A \geq I_n $. Since $ X \geq 0 $, there exists matrix $ D $ such that $ X = DD' $. Further, it follows that \begin{equation} {\rm tr}[AXA'] = {\rm tr}[ADD'A'] = {\rm tr}[D'A'AD] \geq {\rm tr}[D'I_n D] ={\rm tr}[X]. \end{equation}