I am trying to show the following function is convex or not $$f(A)=-\log(\text{trace}(A^{-1}))-\log(\det(A)),$$ where $ A$ is positive definite.
I know $\text{trace}(A^{-1}), -\log(\cdot)$ and $-\log(\det(A))$ is convex. But since $-\log(\cdot)$ is decreasing, simple composition rule for convex function does not work here.
I also tried restrict the $f$ to a line, i.e. let $Z+tV$ where $Z$ positive definite and $Z+tV$ positive definite. Then $$f(Z+tV)=-\log(\text{trace}((Z+tV)^{-1}))-\log(\det(Z+tV)))$$ $$= -\log (\sum_i q_i^T Z^{-1}q_i (1+ t\lambda_i)^{-1}) -\log (Z)-\sum_i\log(1+t\lambda_i),$$
where $\lambda_i$ is the eigenvalue of $Z^{-1/2}VZ^{-1/2}=Q\Lambda Q^T$ and $q_i$ is the $i$th column of $Q$.
Then I compute second derivative (let $a_i=q_i^T Z^{-1}q_i$), which is positive or negative if
$$ -2(\sum \frac{\lambda_i^2a_i}{(1+\lambda_it)^3})(\sum\frac{a_i}{1+\lambda_it})+(\sum \frac{\lambda_ia_i}{(1+\lambda_it)^2})^2+(\sum\frac{\lambda_i^2}{(1+\lambda_it)^2})(\sum\frac{a_i}{1+\lambda_it})^2$$ is positive or negative.
But I do not know how to judge the sign.
Did I miss something or there is a simple way to see this function is convex or not?
If you want to show that $f(A) = - \log({trace( A^{−1} )}) − \log({\det(A)})$ is convex, you have to show that $g(A)= - \log({trace( A^{−1} )})$ is convex.
That is, $g(A)+g(B)-g(\frac{A+B}{2})>0$ $$- \log(trace(A^{−1})) - \log(trace(B^{−1})) + \log(trace(\frac{A+B}{2})^{−1})>0$$ $$\log(\frac{trace(\frac{A+B}{2})^{−1}}{trace(A^{−1})trace(B^{−1})})>0$$ $$\frac{trace(\frac{A+B}{2})^{−1}}{trace(A^{−1})trace(B^{−1})}>1$$ $$trace(\frac{A+B}{2})^{−1}>trace(A^{−1})trace(B^{−1})$$
$$trace(A^{−1})+trace(B^{−1})>2trace(A^{−1})trace(B^{−1}) \ \ \ (1)$$
However, inequality (1) is not true. For a scalar example, $3+1>2\times3\times1$ is not true. Thus, f(A) is not convex from my opinion.