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I'm stuck on an exercise I got in my linear algebra class where you have to find a $5\times5$ matrix $A$ such that $A$ multiplied by itself five times is the identity matrix $I_5.$ $A$ must not be the identity matrix itself.

My problem is that I don't know how to go about this in a clever way. I know that it makes sense to put only ones on the main diagonal of A and only zeroes below the ones but I don't know how to construct the rest of the matrix. It seems tedious to just try out random examples and I want to do it systematically. Any ideas would be appreciated!

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    $\begingroup$ Take the permutation matrix of a 5-cycle in $\mathfrak{S}_5$. E.g. the matrix for $\sigma=(n,1,2,3,...,n-1)$ is a sup-diagonal of ones and a single one last line first column. $\endgroup$ – Anthony Saint-Criq Jan 13 at 17:37
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    $\begingroup$ @P.Quinton Not necessarily. My example still works : different permutations commute iff they have disjoint supports. Take two distinct 5-cycles, their matrix won't commute. $\endgroup$ – Anthony Saint-Criq Jan 13 at 17:41
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    $\begingroup$ Take a matrix that rotates by $2\pi/5$ radians in two dimensions and that leave the other three dimensions fixed. $\endgroup$ – Mankind Jan 13 at 17:44
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    $\begingroup$ @Mankind, I think that that is why @‍mathcounterexamples.net asked about the ground field (or even ring?). If you want a matrix over $\mathbb Z$ or $\mathbb Q$, for example, this rotation isn't good; or, for example, if you're working over $\mathbb C$, then you might as well just take various 5th roots of unity on the diagonal! $\endgroup$ – LSpice Jan 13 at 17:46
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    $\begingroup$ Right, I'm sorry I forgot to mention that it's a matrix over the real numbers! $\endgroup$ – SokraTess Jan 13 at 17:49
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the companion matrix for polynomial $x^5 - 1$ is $$ \left( \begin{array}{ccccc} 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 1&0&0&0&0 \\ \end{array} \right) $$ or, if preferred, the transpose of this.

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A straightforward example would be a matrix that rotates on two of the dimensions by some multiple of $72^\circ$ - as noted in comments here by @BenGrossmann this is a Givens rotation matrix.


... and I now noticed Anthony Saint-Criq's permutation suggestion & subsequent discussion in main comments, that's even easier (as well as being numerically stable), e.g. $\tiny\begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}$ - ensuring you get a 5-cycle as described (not eg 3+2 length cycles, although that would be a neat way to have $A^{\color{violet}{6}}=I_5$ ).

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  • $\begingroup$ Also answered by @Mankind in the comments. $\endgroup$ – LSpice Jan 13 at 18:33
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    $\begingroup$ I think that the idea of rotating "on two of the dimensions" should be clarified. In particular, it is worth noting that the kind of matrix that you have in mind is (I suspect) a Givens rotation. $\endgroup$ – Ben Grossmann Jan 13 at 18:34
  • $\begingroup$ @LSpice if it's worth talking about notes in main comments, it's worth checking the time stamps and seeing that the main comments reference to rotation was only just before my answer and not highlighted at the time. $\endgroup$ – Joffan Jan 13 at 18:56
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    $\begingroup$ Joffan, the permutation matrix is also the companion matrix for $x^5-1$ so little imagination is needed $\endgroup$ – Will Jagy Jan 13 at 19:03
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    $\begingroup$ @WillJagy thanks for another way of viewing this. $\endgroup$ – Joffan Jan 13 at 19:10

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