Assume that M is a n by n matrix, show that there exists M such that $M^5-M-I=0$.
I have been trying to use caley-hamilton thm (and maybe use the characteristic polynomial to find an upper triangular matrix) but since I cannot factor x^5-x-1=0, I do not think I can apply the theorem here. I am a bit stuck on what approach to take here.
The statement in question is not true in general. E.g. it is not solvable when $n=1$ and the underlying field is $GF(2)$. On the other hand, if $n=5$, the equation is satisfied by setting $M$ as the companion matrix for the polynomial $x^5-x-1$, regardless of the underlying field.
If the quintic equation $x^5-x-1=0$ is solvable over the underlying field (it is always solvable if the field is algebraically closed; it is also solvable when the field is $\mathbb R$, because the polynomial has an odd degree), then you don't need companion matrices. You can simply pick a root $\lambda$ of the equation and set $M=\lambda I_n$.