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Is the function $$ \det: A\in \mathbb{P}^{n \times n}(\mathbb{R}) \rightarrow \det (A)$$ a convex function? (where $\mathbb{P}^{n \times n}(\mathbb{R})$ is set of positive definite matrix.)

I think the answer is yes, but I cannot prove it directly using the definition of convex function. How can I do?

Thanks for the help!

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    @angryavian. That counterexample starts with two matrices that aren't positive definite. – 2017-01-25

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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 $\det ({1 \over 2} (A+B)) = {17 \over 8} > 2$.

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    Why the downvote? Please, I need to understand what that downvote is about? – 2017-01-25
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    I upvoted, but I admit that my first thought was like 'hmm, that matrix is not positive definite... oh wait, that is just an orthogonal basis change'. I presume that the downvoter experienced the same confusion. – 2017-01-25
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    @SangchulLee: Thanks! I don't mind downvotes as long as I understand why, but that is difficult to ascertain if the downvoter just does a hit and run. – 2017-01-25