How to express ${\rm Tr}(A^n)$ (in terms of ${\rm det}\,A$), where $A$ is a skew-symmetric $m\times m$ matrix? With references if possible.
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For even powers we cannot determine the trace from the determinant and order alone.
Let matrix $A$ be the $3×3$ matrix defined as follows:
$A_{1,2}=A_{2,3}=A_{3,1}=c,$
with other entries defined by skew-symmetry. Then $\det A=0$. But $A^2$ has all its diagonal elements equal to $-2c^2$, so its trace is $-6c^2$ which varies with $c$ despite the constant determinant.
The trace of an odd power of any skew-symmetric matrix is always zero. Can you see why?
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$\begingroup$ On the vanishing of the trace for odd power, could you explain why? Not sure how to prove it $\endgroup$– KosmMay 5 at 11:13
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1$\begingroup$ $(A^n)^T=-A^n$ for odd $n$, so $A^n$ is also skew-symmetric. $\endgroup$ May 5 at 11:21