Considering in the complex fields. Let $P$ be a nonsingular matrix, $P^\* $ be its conjugate transpose, is there a relation between $P^\*|D|P$ and $|P^\*DP|$, where $D$ is a diagonal matrix? In particular, is it true 


$P^\* |D|P \ge |P^\*DP|$ in the sense of Lowner order, or is there an order for eigenvalues?


Here $|A|=(A^\*A)^{1/2}$, the absolute value of a complex matrix.


**Edit** As I know from Suvrit's answer, there is no relation like $P^\* |D|P \ge |P^\* DP|$ in the sense of Lowner order. So my question becomes, is the $i$th largest eigenvalue of 
$P^\* |D|P$ larger than that of $|P^\*DP|$?



Is there a relation between $P^*|D|P$ and $|P^*DP|$?
**Update:** In the edited question, the OP asks whether $\lambda\_i^\downarrow(P^\*|D|P) \ge \lambda\_i^\downarrow(|P^\*DP|)$ (or even the reverse direction). Such a relations do not hold either. Take for e.g., $P=\begin{bmatrix} 2 & 1 \\\\ 2 & 2\end{bmatrix}$, and use the same $D$ as below. 


Then, we have $\lambda^\downarrow(P^\*|D|P) = (41.22, 0.776)$, while $\lambda^\downarrow(|P^\*DP|) = (23.369, 1.369)$. 


However, if one assume that $P$ is a contraction, then several interesting results can be shown.




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I don't think there is any useful relation.


Here is a counterexample:


\begin{equation\*}
 P =
 \begin{bmatrix}
 2 & 2\\\\
 2 & 4
 \end{bmatrix},
\end{equation\*}
and
\begin{equation\*}
 D = \begin{bmatrix}
 -2 & 0\\\\
 0 & 4
 \end{bmatrix}
\end{equation\*}
Then,
\begin{equation\*}
 P^T|D|P =
 \begin{bmatrix}
 12 & 16\\\\
 16 & 24
 \end{bmatrix}
\end{equation\*}
and


\begin{equation\*}
 |P^TDP|^2 = 
 \begin{bmatrix}
 640 & 1536\\\\
 1536 & 3712
 \end{bmatrix}
\end{equation\*}
Then,
\begin{equation\*}
 \lambda(P^T|D|P - |P^TDP|) = (-3.3678, 31.4856),
\end{equation\*}
which is indefinite.


