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If $A^{n}$ is a symmetric matrix, should I conclude that A is also symmetric?

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    $\begingroup$ Related. $\endgroup$ – Git Gud Jul 30 '15 at 22:22
  • $\begingroup$ Appreciated brothers, its helpful suggestion $\endgroup$ – Patrick Chidzalo Jul 30 '15 at 22:47
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No. Consider the $2 \times 2$ Jordan block $$\pmatrix{0 & 1 \\ 0 & 0},$$ or the matrix $$\pmatrix{0 & -1 \\ 1 & 0},$$ which represents an anticlockwise rotation by $\frac{\pi}{2}$.

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  • $\begingroup$ I appreciate, its helpful answer. $\endgroup$ – Patrick Chidzalo Jul 30 '15 at 22:48
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Willse Jul 30 '15 at 23:57
  • $\begingroup$ More generally, every asymmetric nilpotent matrix is an example, as is every rotation matrix for an angle a rational, nonintegral multiple of $\pi$. $\endgroup$ – Travis Willse Jul 31 '15 at 0:01
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No you can take the nilpotent matrix $$ M=\left(\begin{array}{cccc} 0 & 1 & 0& 0\\ 0&0 & 1& 0\\ 0& 0&0&1\\ 0& 0& 0& 0 \end{array} \right) $$ M isn't symmetrical but $A^4=0_4$ symmetric

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  • $\begingroup$ I appreciate, its helpful answer $\endgroup$ – Patrick Chidzalo Jul 30 '15 at 22:48

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