# $5\times 5$ Matrix $A$ such that $A^5 = I_5$ (in $R^5$)

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!

• 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. – Anthony Saint-Criq Jan 13 at 17:37
• @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. – Anthony Saint-Criq Jan 13 at 17:41
• Take a matrix that rotates by $2\pi/5$ radians in two dimensions and that leave the other three dimensions fixed. – Mankind Jan 13 at 17:44
• @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! – LSpice Jan 13 at 17:46
• Right, I'm sorry I forgot to mention that it's a matrix over the real numbers! – SokraTess Jan 13 at 17:49

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.
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 $$A^{\color{violet}{6}}=I_5$$ ).
• Joffan, the permutation matrix is also the companion matrix for $x^5-1$ so little imagination is needed – Will Jagy Jan 13 at 19:03