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,

  • 1
    $\begingroup$ How do you define the 2-norm? $\endgroup$ – Jonas Dahlbæk Jul 10 '14 at 11:01
  • 1
    $\begingroup$ 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

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 $\|B\|\leq\|A\|$!

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.