Find rotation matrix for multivariate Gaussian such that sum of standard deviations is minimized I'm struggeling with a part of a proof.
Let $A = \mathcal{N}(\mu, \Sigma)$ be a $n-$variate Gaussian, and let $R$ be a $n \times n$ rotation matrix. We can rotate this distribution by the rotation matrix via $\mathcal{N}(R \mu, R\Sigma R^T)$. Now I want to know: under which rotation matrix is the sum of standard deviations minimized?
To formalize, we want to minimize the square rooted elements along the diagonal:
$$\text{argmin}_{R} \sum_{i=1}^n \sqrt{(R\Sigma R^T)_{i,i}}$$

Strong suspicion:
I have a strong suspicion that if we rotate our distribution such that it becomes uncorrelated (use the normalized principle component vectors as a new basis and construct $R$ such that we rotate to that basis), this sum is minimized. This appeared to be the case when I solved this numerically.
In terms of reasoning why, I'm getting stuck.
Steps so far:
The trace (so sum of marginal variances) of $R\Sigma R^T$ remains equal under rotation.
Therefore, the problem "feels" a bit like minimizing $\sum_{i=1}^n |a_i|$ for a set of numbers under the constraint that $C = \sum_{i=1}^n a_i^2$ which is usually done by trying to maximize the difference between the $a_i$'s, but that's how far I got.
Maybe we can use the fact that the uncorrelated basis/principle component basis is used in PCA as it is the direction that explains the maximum amount of variance?
Another way I tried to look at it is by taking an uncorrelated Gaussian and showing that any rotation would increase the sum of marginal standard deviations, but that didn't help much either.
 A: Consider any $n\times n$ real symmetric PSD matrix $B=UDU^T$ where $U$ is orthogonal and $D$ is diagonal.  Now collect the diagonal elements of $B$ in vector $\mathbf b$ and collect the eigenvalues of $B$ (diagonals of $D$) in vector $\mathbf d$.
1.) $\mathbf b\preceq \mathbf d$
which reads: $\mathbf b$ is (strongly) majorized by $\mathbf d$
this is a corollary of
Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$
with the additional constraint that the columns of $Q$ are standard basis vectors which proves that for any $m\in \big\{1,2,\dots,n\big\}$
$\sum_{k=1}^m b_{[k]} \leq \sum_{k=1}^m d_{[k]}$
(where e.g. $b_{[k]}$ denotes the kth largest value in $\mathbf b$)
And recall for the case of $m=n$ that
$\sum_{k=1}^n d_{k}=\text{trace}\big(D\big)=\text{trace}\big(UDU^T\big)=\text{trace}\big(B\big)=\sum_{k=1}^n b_{k}$
2.) for $x\geq 0$ note that $x\mapsto x^\frac{1}{2}$ is concave (check 2nd derivative), thus $f:\mathbb R_{\geq 0}^n\longrightarrow \mathbb R$ given by
$f\big(\mathbf a\big)= \sum_{k=1}^n a_k^\frac{1}{2}$ is Schur Concave
Putting (1) and (2) together gives
$\text{trace}\big((B\circ I)^\frac{1}{2}\big)=f\big(\mathbf b\big)\geq f\big(\mathbf d\big) = \text{trace}\big(D^\frac{1}{2}\big)$
