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The determinant of a matrix in $\mathfrak h_3(\Bbb O)$ is defined by

$(a,b,c,\mathbf{a},\mathbf{b},\mathbf{c})=\begin{bmatrix} a &\mathbf{c} &\mathbf{b} \\ \mathbf{\bar{c}} & b & \mathbf{a}\\ \mathbf{\bar{b}}& \mathbf{\bar{a}} & c \end{bmatrix}$

$a,b,c \in \mathbb{R} $ and $\mathbf{a},\mathbf{b},\mathbf{c}\in \mathbb{O}$

$det(a,b,c,\mathbf{a},\mathbf{b},\mathbf{c})=abc-(a\left \| \mathbf{a} \right \|^{2}+b\left \| \mathbf{b} \right \|^{2}+c\left \| \mathbf{c} \right \|^{2})+2Re(\mathbf{abc}) $

How we can express this in terms of the trace?

$det(x)=\frac{1}{3}tr(x^{3}) - \frac{1}{2}tr(x^{2}))tr(x)+ \frac{1}{6}tr(x)^{3} for x\in \mathfrak{h}_{3}(\mathbb{O}(\mathbb{Z}_{q}))$

My question is

How can we express determinant in terms of trace?

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    $\begingroup$ en.wikipedia.org/wiki/Cayley-Hamilton_theorem. Use the Hamilton-Cayley theorem (Blue-highlightened formula) $\endgroup$ – guestDiego Oct 13 '16 at 13:31
  • $\begingroup$ @guestDiego Cayley-Hamilton wouldn't seem to apply, since $\Bbb O$ is neither commutative nor associative. $\endgroup$ – arctic tern Mar 28 '17 at 22:18

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