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Given a fixed $B \in \mathfrak{su}(4)$ is it possible to solve for $F$:

$\sigma^{\text{max}}\left(\frac{A}{F(A)} + B \right) = 1$, $\forall A \in \mathfrak{su}(4)$. A theorem in http://arxiv.org/abs/math/0109060 ensures that $F$ will be a (potentially non reversible) norm. Here $\sigma^{\text{max}}$ is the largest singular value of its argument.

I am particularly interested in the case when $B = i \begin{pmatrix} 1/6 & 0 & 0 & 0 \\ 0 & -(1/6) & 1/3 & 0 \\ 0 & 1/3 & -(1/6) & 0 \\ 0 & 0 & 0 & 1/6 \end{pmatrix}$.

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  • $\begingroup$ Your particular $B$ is not in ${\frak{su}}(4)$, which consists of traceless skew-Hermitian matrices. Do you mean your example to be $iB$? $\endgroup$
    – Robert Bryant
    Commented Sep 10, 2014 at 9:53
  • $\begingroup$ I've found the form of $F$ on the subspace for which $[A,B]=0$. I may have found a method that treats the subspace $Tr(AB)=0$ also but I'm less sure. $\endgroup$
    – Benjamin
    Commented Sep 10, 2014 at 10:29

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You want a positive (I assume) number $1/s$ such that $\det(sA + B \pm iI) = 0$.
For any given $A$ and choice of $\pm$, that is a polynomial in $s$ of degree at most $4$. So yes, it can be solved in ``closed form", but it's not likely to be pretty.

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  • $\begingroup$ For $A=0$ we have $\det(B\pm iI)\neq 0$. $\endgroup$
    – Dietrich Burde
    Commented Dec 10, 2014 at 10:29
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    $\begingroup$ Obviously $A=0$ might not work. The "solve for $F$" should be interpreted as "find $F$ if it exists, otherwise output 'does not exist'". But $\det(B \pm iI)$ could be $0$, e.g. $B$ could be diagonal with some entries $\pm i$. $\endgroup$
    – Robert Israel
    Commented Dec 10, 2014 at 22:50

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