While reading a paper An Arithmetic Proof of John’s Ellipsoid Theorem by Gruber and Schuster, I have a question on their proof.
Consider an $n\times n$ real symmetric and positive definite matrix $\mathbf A$.
- As this kind of matrix is symmetric, its $n(n+1)/2$ upper diagonal terms are enough to represent it. Hence, we can consider such a matrix as a point in $\mathbb R^{n(n+1)/2}$.
- A conical combination of two positive definite matrices is also positive definite. Hence, the set of all symmetric positive definite matrices forms an open convex cone $\mathcal P\in\mathbb R^{n(n+1)/2}$ with apex on the origin.
Now they claim the following theorem without proving it.
Theorem: The set $\ \mathcal D = \{\mathbf A \in \mathcal P: \det \mathbf A \geq 1\}$ is a closed, smooth, strictly convex set in $\mathcal P$ with non-empty interior.
They just gave some hint that we can use Implicit Function Theorem and Minkowski's Determinant Inequality which states that
For two $n\times n$ positive semidefinite Hermitian matrices $\mathbf X$ and $\mathbf Y$,
$$\det (\mathbf X + \mathbf Y)^{1/n}\geq \det(\mathbf X)^{1/n} + \det(\mathbf Y)^{1/n} $$
Any hint or suggestion on how to prove the above theorem about the set $\mathcal D$?
Best Answer
Here is a textbook level description of the above. I assume you know what a convex set and convex function on this set are. Given that, let us know prove that the determinant is strictly log-concave on hermitian positive definite matrices.
Claim. Let $A, B > 0$. Then, $\det\left(\frac{A+B}{2}\right) \ge \sqrt{\det(AB)}$
Proof Consider $\phi(A) := \log\det(A)$. The first derivative of this is $A^{-1}$, while the second derivative may be identified with $-A^{-1}\otimes A^{-1}$, which is clearly negative definite if $A > 0$. This proves the desired concavity of $\phi(A)$, and therewith the claim above.
Note: Minkowski's determinant inequality is not the same as the above log-concavity. It is stronger, and enjoys a variety of different proof attempts. For a great list of these, have a look at the following much older MO question.