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Let $A$ be an $n$ by $n$ real matrix of order $d$. i.e. $d$ is the smallest positive integer greater than $1$ that makes $A^{d}=I_{n}$. The set of trace zero real matrices form $n^{2}-1$ dimensional real vector space. Let us call this vector space $V$. Let $\phi_{A}$ be the map from $V$ to $V$ sending $X$ to $AXA^{-1}$. Then $\phi$ is a linear transformation of $V$. I would like to know that if there is known relationship between the trace of $A$ and the trace of $\phi_{A}$.

Since $A$ is of finite order $A$ is conjugate to the direct sum of $2$ by $2$ rotation matrices $B_{\theta_{i}}$s, some identity matrix $I_{k_{1}}$ and some $-$identity matrix $-I_{k_{2}}$. Here the rotation matrix $B_{\theta_{i}}$ is the matrix \begin{displaymath} \left( \begin{array}{cc} \cos \theta_{i} & -\sin \theta_{i} \ \sin \theta_{i} & \cos \theta_{i} \

\end{array} \right) \end{displaymath} where $\theta_{i}$ is a rational multiple of $2\pi$.

Since $\phi_{A}$ is a linear transormation of finite order, there exists basis of $V$ such that the matrix for $\phi_{A}$ is a form of direct sum of rotation, Identity and $-$identity matrices. But is there any relation between trace of $A$ and trace of $\phi_{A}$ in terms of the angles $\theta_{i}s$?

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    $\begingroup$ For any $n \times n$ matrices $A$, $B$, the trace of the map $X \in M_n \mapsto A X B \in M_n$ is $Tr(A) Tr(B)$. Here you therefore get $Tr(\phi_A) = Tr(A) Tr(A^{-1}) - 1$. $\endgroup$ Jul 26, 2012 at 14:54
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    $\begingroup$ This is not at an appropriate level for MO (see the FAQ). But see math.stackexchange.com. $\endgroup$ Jul 26, 2012 at 14:57

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You have a (real) linear representation r of the cyclic group C of order d; you give its character, and then ask for the character of a certain subrepresentation of the tensor product of r and its dual? Well, at the level of characters this is not a deep question. The subrepresentation is the complement of the subspace of multiples of the identity matrix.

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