# Log determinant of quadratic form

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.

• @RodrigodeAzevedo: yes, these are quadratic forms. I just get those directly from the paper. Commented Feb 15, 2022 at 11:09
• A quadratic form is a polynomial where each monomial is of degree $2$. Commented Feb 15, 2022 at 11:09
• If the quadratic forms are positive definite, do you really need to take the absolute value? Commented Feb 15, 2022 at 11:11
• 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). Commented Apr 15, 2022 at 13:43

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.