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

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

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.

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