## Is there a relation between $P^*|D|P$ and $|P^*DP|$?

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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 $|P^*DP|$?

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Even the modified question has a negative answer; please see my update. – S. Sra Oct 26 2011 at 10:42

## 2 Answers

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.

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.

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Isn't your $P^TDP=[8, 24; 24, 56]$, then $|P^TDP|=[11.3137, 22.6274; 22.6274, 56.56858]$? – Sunni Jul 12 2011 at 23:54
sorry, while typesetting I reran with 'sqrt' instead of 'sqrtm' by mistake; it's still a counterexample with the matrix squareroot--- thanks for catching. – S. Sra Jul 13 2011 at 1:54
I think there is a relation between eigenvalues, i.e. the $k$th largest eigenvlaue of $P^* |D|P$ is no less than the $k$th largest eigenvalue of \$|P^*DP| – Zae Kwong Jul 13 2011 at 17:12
Sure, one can have several inequalities; All I am saying is that the semidefinite ordering that you asked in your second question above, does not hold in either direction because of the above indefinite matrix. – S. Sra Jul 13 2011 at 17:40

1
Is there any reason why you do not divide everything by 2? – Tsuyoshi Ito Jul 22 2011 at 13:40
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Using the same example above, I found

p=[1,1;1,2]

p =

```` 1     1
 1     2
````

d=[-1,0;0,2]

d =

````-1     0
 0     2
````

c=[1,0;0,2]

c =

```` 1     0
 0     2
````

eig(((p*d*p)^2)^(1/2))

ans =

````0.2426
8.2426
````

eig(p*c*p)

ans =

````0.1690
````

11.8310

But this example shows it is possible the majorization relation holds.

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