Set of Positive Definite matrices with determinant > 1 forms a convex set

Question

While reading a paper An Arithmetic Proof of John’s Ellipsoid Theorem by Gruber and Schuster, I have a question on their proof.

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Consider an $n\times n$ real symmetric and positive definite matrix $\mathbf A$.

1. 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 $\mathbb R^{n(n+1)/2}$.
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.

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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 $\mathcal D$?

Answer 1

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.

Answer 2

$-\log \det$ is a smooth convex function on the PSD cone (this is a standard fact , and follows from the Chandler Davis theorem -- see, e.g., my arXiv preprint on "another proof of the Davis theorem", or see Boyd and Vanderberghe's convex optimization for deep significance of this fact in convex programming), so the set $\log \det X > a$ is a smooth convex set.

Answer 3

I'm not sure if you are asking for the proof of the theorem assuming Minkowski's inequality or if you are asking for the proof including a proof of Minkowski's inequality.

If you assume Minkowski's inequality there is nothing to prove: the sup-level sets of a concave function are convex. In this case, if you take two positive definite Hermitian matrices $A$ and $B$ with determinant greater or equal to $1$, then $$ \det((A + B)/2)^{1/n} \geq (\det(A)^{1/n} + \det(B)^{1/n})/2 \geq 1 . $$

Now, the proof of Minkowski's inequality is just the arithmetic-geometric mean inequality once you follows Misha's line of thought in the comments to the OP.