While reading a paper [*An Arithmetic Proof of John’s Ellipsoid Theorem*](http://dmg.tuwien.ac.at/schuster/john1.pdf) 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](http://en.wikipedia.org/wiki/Positive-definite_matrix) 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](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/Implicit_function_theorem) and Minkowski's Determinant Inequality which states that



> 
> For two $n\times n$ positive semidefinite [Hermitian](http://en.wikipedia.org/wiki/Hermitian_matrix) 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 $\mathcal D$?



Set of Positive Definite matrices with determinant > 1 forms a convex set
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](https://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n).


