This is (probably) an easy one:


Given a positive definite matrix $M$, find the positive definite matrix $X$, which minimizes $\textrm{tr}(X M)$ subject to $\det(X) = 1$.


Looking for how to find X, either a formula, an algorithm, or a reference to a paper. 


-Thanks



Matrix optimization problem
If the eigenvalues of $M$ are $\lambda\_1$, $\lambda\_2$, ..., $\lambda\_n$, then the minimum is $n (\lambda\_1 \cdots \lambda\_n)^{1/n}$. 


**Proof:** Your problem is invariant under conjugating $X$ and $M$ by a unitary matrix. So we may assume that $M$ is diagonal, with diagonal entries $\lambda\_1$, ..., $\lambda\_n$. Once we have made this assumption, we have $\mathrm{Tr}(XM) = \sum x\_{ii} \lambda\_i$. In other words, this is a linear function of the diagonal of $X$.


We temporarily restrict ourselves to looking at those $X$ whose eigenvalues are $\mu\_1$, $\mu\_2$, ..., $\mu\_n$. The space of positive definite matrices $X$ with eigenvalues $\mu\_1$, $\mu\_2$, ..., $\mu\_n$ is a compact manifold. The [Schur-Horn theorem](http://en.wikipedia.org/wiki/Schur%25E2%2580%2593Horn_theorem) states:



> 
> The subset of $\mathbb{R}^n$ which can occur as the diagonal of a positive definite matrix with eigenvalues $(\lambda\_1, \ldots, \lambda\_n)$ is a convex polytope;
>  its vertices are the $n!$ permutations
>  of $(\mu\_1, \ldots, \mu\_n)$.
> 
> 
> 


A linear functional on a convex polytope is always minimized at a vertex. We conclude that



> 
> Given $(\mu\_1, \ldots, \mu\_n)$, as
>  $X$ ranges over positive definite
>  matrices with eigenvalues $\mu\_i$, the
>  minimal value of $\mathrm{Tr}(XM)$ is
>  $\sum \lambda\_i \mu\_i$, where the
>  $\lambda$'s and the $\mu$'s are sorted
>  in opposite orders.
> 
> 
> 


Now, we want to vary the $\mu$'s. So we want to find the minimal value of $\sum \lambda\_i \mu\_i$ where $\mu\_i$ ranges over $n$-tuples obeying $\prod \mu\_i = 1$, with $\mu\_i$ sorted in the reverse order from $\lambda\_i$. (In fact, removing this last condition will not effect the minimum.) 


By the AM-GM inequality, $\sum \lambda\_i \mu\_i \geq n \prod \lambda\_i^{1/n} \prod \mu\_i^{1/n} = n \prod \lambda\_i^{1/n}$. We get equality if $\mu\_i = \left( \prod \lambda\_i \right)^{1/n} / \lambda\_i$, and we take $X$ to be diagonal in the same basis where $M$ is diagonal.


