# Norm of symmetric positive semidefinite matrices

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 ?

Regards.

• How do you define the 2-norm? – Jonas Dahlbæk Jul 10 '14 at 11:01
• 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. – Hussein Hammoud Jul 10 '14 at 11:02
• Wouldn't that coincide with the operator norm? – 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 $\|B\|\leq\|A\|$!
• $C$ is positive definite if and only if $C$ is positive semi-definite and invertible. In this case, the lowest eigenvalue of $C$ is $\|C^{-1}\|^{-1}$. – Jonas Dahlbæk Jul 11 '14 at 9:54