Among all $n\times n$PSD matrices $C$ with eigenspectrum $\lambda_1\ge \dots\ge \lambda_n \ge 0,$ which one achieves maximum/minimum expected norm $E_{w\sim N(0,C)} \|w\|$?
In the case of squared norm we can rewrite the vector in terms of a normalized Gaussian
\begin{align}
E \|w\|^2 &= E tr(w^\top w) \\
&= E tr(g^\top C g) && g\sim N(0,I_n) \\
&= tr(C) = \sum_i^n \lambda_i
\end{align}
Therefore, given the eigenspectrum of $C$, the choice of eigenvectors of $C$ does not affect the expected squared norm of the Gaussian vector.
The same does not seem to hold for $E \|w\|$. I wonder if the case of diagonal matrix $C = diag(\lambda_1,\dots,\lambda_n)$ holds the min/max expected norm $E \|w\|$?
If $X\sim N(0,C)$ let us show that $E(\|X\|)$ depends on the eigenvalues $(\lambda_1,\ldots,\lambda_n)$ of $C$ only. As usual $X\sim C^{1/2}Z$ with $Z\sim N(0,I_n).$ The trick is to use the following integral representation of the function $$x\mapsto \sqrt{x}=\frac{2}{\sqrt{\pi}}\int_0^{\infty}\frac{1-e^{-tx}}{t^{3/2}}dt$$ which can be checked by derivation with respect to $x$. We apply it to $$x=\|X\|^2\sim \lambda_1Z_1^2+\cdots+\lambda_nZ_n^2.$$ Recall that $Z_1,\cdots,Z_n$ are independent. Therefore the expectation of $\|X\|$ is $$E(\|X\|)=\frac{2}{\sqrt{\pi}}\int_0^{\infty}\frac{1-E\left(e^{-t(\lambda_1Z_1^2+\cdots+\lambda_nZ_n^2)}\right)}{t^{3/2}}dt=\frac{2}{\sqrt{\pi}}\int_0^{\infty}\left(1-\frac{1}{\prod_{i=1}^n\sqrt{1+2\lambda_i t}}\right)\frac{dt}{t^{3/2}}.$$