0
$\begingroup$

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 $B = i \begin{pmatrix} 1/6 & 0 & 0 & 0 \\ 0 & -(1/6) & 1/3 & 0 \\ 0 & 1/3 & -(1/6) & 0 \\ 0 & 0 & 0 & 1/6 \end{pmatrix}$.

$\endgroup$
2
  • $\begingroup$ Your particular $B$ is not in ${\frak{su}}(4)$, which consists of traceless skew-Hermitian matrices. Do you mean your example to be $iB$? $\endgroup$ 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

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ For $A=0$ we have $\det(B\pm iI)\neq 0$. $\endgroup$ Commented Dec 10, 2014 at 10:29
  • 1
    $\begingroup$ 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 $\pm i$. $\endgroup$ Commented Dec 10, 2014 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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