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I have been researching a lot trying to find an answer to my question and didn't find any so I would appreciate it if anyone can help.
If we have 2 symmetric, positive semi-definite matrices $A$ and $B$ for which they satisfy:
$2-Norm(A) \ge 2-Norm(B)$.
How can we relate the 2-Norms of $CAC$ and $CBC$ where $C$ is a Toeplitz, symmetric and also positive semi-definite matrix ?

Thank you for your help,
Regards.

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    $\begingroup$ How do you define the 2-norm? $\endgroup$ – Jonas Dahlbæk Jul 10 '14 at 11:01
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    $\begingroup$ the 2-norm of a matrix is the square root of the largest singular value of it. Since A is symmetric, the 2-norm of A is simply the largest eigenvalue of A. $\endgroup$ – Hussein Hammoud Jul 10 '14 at 11:02
  • $\begingroup$ Wouldn't that coincide with the operator norm? $\endgroup$ – Jonas Dahlbæk Jul 10 '14 at 11:10
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Observe that the $2$-norm of a matrix is $$\|A\|_2=\sup_{\|x\|=1}\sqrt{\langle x,A^*Ax\rangle}=\sup_{\|x\|=1}\sqrt{\|Ax\|^2}=\|A\|,$$ i.e. the 2-norm and the operator norm coincide. Knowing only $\|B\|\leq\|A\|$, I would not expect to be able to control $\|CBC\|$ in terms of $|\|CAC\|$. The following example shows that you need some type of extra condition to get a bound: $$ A=\left(\matrix{ 1&-1\\ -1&1 }\right), C=\left(\matrix{ 1&1\\ 1&1 }\right). $$ For these matrices, you get $CAC=0$.

Case 1: If we assume that $C$ is invertible, then $$ \|CAC\|\geq \|C^{-1}\|^{-1}\|AC\|=\|C^{-1}\|^{-1}\|(AC)^*\|=\|C^{-1}\|^{-1}\|CA\| \geq \|C^{-1}\|^{-2}\|A\|, $$ so $$ \|CBC\|\leq \|C\|^2\|B\|\leq\|C\|^2\|A\|\leq\|C\|^2\|C^{-1}\|^2\|CAC\|. $$

Case 2: Suppose $A$ is invertible, such that $I\leq \|A^{-1}\| A$. Observe that $B\leq\|B\|I\leq \|B\|\|A^{-1}\|A$, which implies $CBC\leq \|B\|\|A^{-1}\|CAC$, and thus $$ \|CBC\|\leq \|B\|\|A^{-1}\|\|CAC\|. $$ Note that this holds even without the condition $\|B\|\leq\|A\|$!

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  • $\begingroup$ What if we add the condition that C is strictly positive definite (Instead of positive semi-definite) ? Do we get any better bounds/results ? $\endgroup$ – Hussein Hammoud Jul 11 '14 at 9:42
  • $\begingroup$ $C$ is positive definite if and only if $C$ is positive semi-definite and invertible. In this case, the lowest eigenvalue of $C$ is $\|C^{-1}\|^{-1}$. $\endgroup$ – Jonas Dahlbæk Jul 11 '14 at 9:54

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