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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.

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  • $\begingroup$ @RodrigodeAzevedo: yes, these are quadratic forms. I just get those directly from the paper. $\endgroup$
    – lmn2609
    Commented Feb 15, 2022 at 11:09
  • $\begingroup$ A quadratic form is a polynomial where each monomial is of degree $2$. $\endgroup$
    – Rodrigo de Azevedo
    Commented Feb 15, 2022 at 11:09
  • $\begingroup$ If the quadratic forms are positive definite, do you really need to take the absolute value? $\endgroup$
    – Rodrigo de Azevedo
    Commented Feb 15, 2022 at 11:11
  • $\begingroup$ log is an increasing function, we may safely remove it from both sides. Then, since $B^\top \Sigma^{-1} B\leqslant B^\top \Delta^{-1} B$, determinant of LHS does not exceed determinant of right hand side (since corresponding eigenvalues on the left do not exceed these on the right). $\endgroup$
    – Fedor Petrov
    Commented Apr 15, 2022 at 13:43

1 Answer 1

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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.

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