I am reading a paper Cook and Forzani – Likelihood-Based Sufficient Dimension Reduction where the author uses the following result from matrix analysis but does not explain why it is true nor provide any reference.
More specifically, let $B \in \mathbb{R}^{p\times d}$ be a semi-orthogonal matrix, i.e $B^\top B = I_d$, and $d < p$. Let $\Sigma$ and $\Delta$ denote two symmetric positive definite matrices such that $\Sigma – \Delta$ is also positive definite. What they claim is that
$$ \log \det \left\lvert B^\top \Sigma^{-1} B \right\rvert \leq \log \det \left\lvert B^\top \Delta^{-1} B \right\rvert.$$
Could someone point me in the direction of explaining why it is true?
I am thinking of using the Poincaré separation theorem, which provides bounds on the eigenvalues of the matrices on both the left and right-hand sides; however, I did not make any progress with it.
The claim is in the proof of Proposition 3, at the top of page 33. Here $\Sigma = \operatorname{Var}(X)$, and $\Delta = E (\operatorname{Var}(X\vert y))$, so $\Sigma-\Delta = \operatorname{Var}(E(X \vert y))$ is also positive definite. The $B$ in my question plays the role of $B_0$ in the paper.
Best Answer
Since $\Sigma \succ \Delta$, by operator monotonicity we have $\Sigma^{-1} \prec \Delta^{-1}$ and thus $B^{\top}\Sigma^{-1}B \prec B^{\top}\Delta^{-1}B$. Since log on positive real is increasing, the trace monotonicity (also Theorem 2.10 here) gives $\text{trace}\left(\log\left(B^{\top}\Sigma^{-1}B\right)\right) < \text{trace}\left(\log\left(B^{\top}\Delta^{-1}B\right)\right)$ where the log inside the last inequality is principal logarithm. Replacing $\text{trace}\left(\log\left(\cdot\right)\right)$ with $\log\det(\cdot)$, we get the desired inequality.