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We know that $\log \det (X)$ is a concave function if $X$ is a positive definite matrix. Furthermore, we know that $\det (X^{-1})=(\det(X))^{-1}$ which means that $\log \det(X^{-1})$ is a convex function if $X$ is positive definite. However, we know that the inverse of a positive definite matrix is a positive definite matrix and therefore $\log \det(X^{-1})$ should also be a concave function for $X$ being a positive definite matrix. Where am I wrong in this reasoning?

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    $\begingroup$ Please learn to write an informative (not generic) title, so others can find the problem and solution later. $\endgroup$ Aug 13, 2019 at 18:36
  • $\begingroup$ @NicNic8 I mean if the domain of this function is constrained to the positive definite matrices only then its a concave function. Surely if we fix the input then its a constant. $\endgroup$ Aug 13, 2019 at 18:38
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    $\begingroup$ @NicNic8: You are mistaken, $\log\det X$ is indeed concave in $X$. $\endgroup$
    – user856
    Aug 13, 2019 at 18:40
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    $\begingroup$ @Frank: How do you justify your last argument that $\log\det X^{-1}$ should be concave? Have you tried an analogous argument for $\log x^{-1}$ on the positive reals? $\endgroup$
    – user856
    Aug 13, 2019 at 18:45

2 Answers 2

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Your first argument is correct, for as you noted when $X\succ 0$, $$ \log\det(X^{-1})=\log((\det(X))^{-1})=-\log\det(X), $$ which is the negative of a concave function, hence convex.

Your issue with the second argument is you seem to be using the "claim" that if $f:C\to \mathbb{R}$ is concave, where $C$ is a convex set, then for any invertible map $\phi:C\to C$, $f\circ \phi$ is still concave because it has the same domain. But that clearly is not true: consider $f:\mathbb{R}_{>0}\to \mathbb{R}$ given by $f(x)=x$, and $\phi:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$ given by $\phi(x)=1/x$. $f$ is concave on this domain, while of course $(f\circ \phi)(x)=1/x$ is well-defined and convex on the same domain, but your reasoning would suggest that it should be concave as well.

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That “therefore” makes no sense. See what happens with the real numbers greater than $0$: $\log(x)$ is concave and $\log\left( x^{-1}\right)=\bigl(-\log(x)\bigr)$ is convex, not concave.

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