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Let $M_n(C)$ denote the $n\times n$ matrices with complex entries acting on the Hilbert space $C^n$. As norm on $M_n(C)$ we take the operator norm, i.e. the largest eigenvalue of its absolute value. Now let $\Phi:M_n(C)\rightarrow M_n(C)$ be an explicitly known linear operator. How can we numerically compute the operator norm of $\Phi$?

So far I have tried to use random sampling (taking random matrices and comparing the norms before and after), but this does not scale well with $n$, and formulating it as an optimisation problem in Mathematica, but this only gave results in the $2\times 2$ case.

PS: This is my first question here, please let me know if I should add something.

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  • $\begingroup$ Strange. Is this a question generated by KI? (eigenvalue of its absolute value, Hilbert space $C^n$,...) $\endgroup$
    – Dieter Kadelka
    Commented Oct 7, 2022 at 13:10
  • $\begingroup$ Do you have to implement this yourself and does it have to be in Mathematica? The classic approach would be to compute the singular value decomposition for which there's a good implementation in scipy. $\endgroup$
    – Daniel Shapero
    Commented Oct 8, 2022 at 2:13
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    $\begingroup$ The problem is the norm that you are considering. Taking the largest singular value gives the norm of a matrix acting on the Hilbert space $C^n$, i.e. the norm of the element of the C*-algebra $M_n(C)$. However, what I want to know is the norm of $\Phi$, which is an operator from the C*-algebra to itself. Therefore, the norm should not be given by the singular value decomposition, if I am not mistaken. I would be fine with any way of doing it, and if it is already implemented this would be best. I also do not insist on Mathematica. $\endgroup$
    – Matthijs
    Commented Oct 11, 2022 at 9:49

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