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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?

Rodrigo de Azevedo
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topgrace
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2 Answers2

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

copper.hat
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  • Why the downvote? Please, I need to understand what that downvote is about? – copper.hat Jan 25 '17 at 04:00
  • 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. – Sangchul Lee Jan 25 '17 at 04:02
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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. – copper.hat Jan 25 '17 at 04:03
  • @copper.hat so it is concave? – Turbo Jul 19 '19 at 03:23
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    @Turbo: The function $\log \circ \det$ is concave and $x \mapsto e^x$ increasing, so yes, it is concave. (See https://math.stackexchange.com/q/1192329/27978.) – copper.hat Jul 19 '19 at 20:20
  • @Turbo: Actually, that would be an easy proof. Since it is concave, then if it was also convex it would be linear (affine, and zero at zero) which it is not for $n >1$. – copper.hat Jul 20 '19 at 22:07
  • @copper.hat Just sincerely ask you that why not just say $\log$ is nondecreasing (you say $e^x$ is increasing)? I know you want to say since $\log$ nondecreasing and concave and $\log\circ\det$ concave so $\det$ concave by composition rule. – sleeve chen Jul 28 '19 at 09:47
  • @sleevechen: Sorry, I am not sure what you are asking. It is a fairly well known proof that $\log \circ \det$ is concave. – copper.hat Jul 29 '19 at 11:43
  • @copper.hat I mean why you mention $x\mapsto e^x$ is increasing instead of $\log$ is nondecreasing. (please see the comment at the time 20:20) – sleeve chen Jul 29 '19 at 18:03
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    @sleevechen: It is well known that $\log \circ \det$ is concave, and the composition of increasing with concave is concave, hence $\det$ (as above) is concave. – copper.hat Jul 29 '19 at 20:01
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    @copper.hat $x->x$ is concave, $x->e^x$ not concave, it is convex. increasing function does not generally preserve convexity or concavity. – user138668 Jul 31 '19 at 07:28
  • Why the downvote? Just because? – copper.hat May 04 '21 at 18:57
  • Downvote because you did not give a straight answer. – mxdxzxyjzx Jun 29 '21 at 20:28
  • @copper.hat “no, determinant is not convex.” Then why. – mxdxzxyjzx Jun 29 '21 at 22:15
  • @mxdxzxyjzx Do you understand what a counterexample is? – copper.hat Jun 29 '21 at 22:26
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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

F_G
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