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


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.


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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