Let a $\mathbb{Q}$-vector space $V$ have a $\mathbb{Q}$-linear endomorphism $T : V\rightarrow V$ such that $T^4+14T+28=0$. Prove that $4|\dim{V}$ and for $\dim{V}=4$, $T$ is unique up to similarity.
I am unsure how to approach this. I have tried using minimal polynomial divisibility to no avail. Any ideas on how to approach this?
Observe that $\;x^4+14x+28\in\Bbb Q[x]\;$ is irreducible (why?) , so this one is the minimal polynomial of $\;T\;$ over the linear space $\;V_{\Bbb Q}\;$ . But this minimal polynomial divides the characteristic polynomial of $\;T\;$ , which has degree equal to $\;\dim V\;$ ...