Let $\mathbb{P}^{n \times n}(\mathbb{R})$ denote the set of positive definite matrices. Is the following function convex? $$ \det: A\in \mathbb{P}^{n \times n}(\mathbb{R}) \to \det (A)$$
I think the answer is yes, but I cannot prove it directly using the definition of convex function. How can I do?
On
The determinant is not convex, nor concave, as soon as $n \ge 2$.
If a function is convex, it is convex when restricted to a subspace. If the determinant were convex, it would be convex on the set of (positive definite) diagonal matrices. But then, in the case $n=2$ for example, the determinant of $\begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix}$ is $x y$ which is not convex (indefinite Hessian). Same holds for any $n \ge 2$.
The same reasoning shows that the determinant is not concave as soon as $n \ge 2$.
However it is known the $n$-th root of the determinant is concave on positive definite matrices, see e.g. https://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n
Let $Q={1 \over \sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$, $A=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$, $B= QAQ^T$.
Then $A>0,B>0$ and $\det A = \det B = 2$, but $\det ({1 \over 2} (A+B)) = {17 \over 8} > 2$.