Given a $p \times p$ positive definite matrix $\Sigma$, why eigenvectors of $\Sigma$, stacked as columns of a matrix $R \equiv [r\_1 \, r\_2 \, \ldots \, r\_p]$, optimize the following orthogonally constrained minimization problems?
$$
\mathrm{minimize}~~~~\log \det\big( I \odot (R^\mathsf{T}\Sigma R) \big)~~~~~\text{s.t.}~~~~~R^\mathsf{T}R = I~~,
$$
where $\odot$ is Hadamard product, or equivalently,
$$
\mathrm{minimize}~~~~\sum\_{i=1}^p \log(r\_i^\mathsf{T}\Sigma r\_i)~~~~~\text{s.t.}~~~~~r\_i^\mathsf{T}r\_j = \delta\_{ij}~~.
$$



Why eigenvectors optimize this orthogonally constrained nonlinear minimization problem?
Your minimization problem is equivalent to
\begin{equation\*}
\min\_{R^TR=I}\quad\prod\_{i=1}^p r\_i^T\Sigma r\_i,
\end{equation\*}
and it can be shown (using *Hadamard's determinant inequality* and some more argumentation) that this minimum overall $p$ orthonormal tuples is achieved by choosing the $r\_i$ corresponding to the smallest $p$ eigenvectors.


**EDIT** Here the details that complete my answer (because it seems that the OP already knew the answer and still posted this question...)


We know that for a semidefinite matrix $\Sigma$ with eigenvalues $\lambda\_1,\ldots,\lambda\_n$, and any orthonormal $R$,


\begin{equation\*}
 \prod\_{i=n-p+1}^n \lambda\_i \le \det(R^T\Sigma R) \le \prod\_{i=1}^p (R^T\Sigma R)\_{ii} = \prod\_{i=1}^p r\_i^T\Sigma r\_i.
\end{equation\*}
We are trying to minimize the upper, bound and by choosing $R$ to be the matrix of the smallest $p$ eigenvectors, we can actually turn the first inequality into an equality. This proves the claim. The first inequality is a classic result on eigenvalues, while the second inequality is *Hadamard's determinant theorem*.


