Question
Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and suppose the linear operator $T$ satisfies $T^4-1=T^3+3T^2+T+3=0$. Show $V$ is the direct sum of two dimensional $T$-invariant subspaces.
Attempt
Factoring shows that $T$ is annihilated by both $p(t)=(t-1)(t^2+1)(t+1)$ and $g(t)=(t+3)(t^2+1)$. The minimum polynomial $\mu$ must divide both $p(t)$ and $g(t)$ so we see that $\mu=(t^2+1)$. Since the characteristic polynomial $\chi$ of $T$ and $\mu$ have the same roots, $\chi(t)=(t^2+1)^n$. I am unsure how to reason further. Perhaps: Since the rational canonical form exists, there is a direct decomposition of $V$ into cyclic subspaces of $V$, i.e $$V=C_{x_1} \oplus \ldots \oplus C_{x_k}.$$ Note that $C_{x_i}$ is $T$ invariant. Thus the minimum polynomial of $T\mid_{C_{x_k}}$ divides $\mu$ and since $t^2+1$ does not split over $\mathbb{R}$, it must be the case that the minimum polynomial of the restriction of $T$ to each of these subspaces is also $t^2+1$ This implies that each cyclic subspace is $2$ dimensional (I am actually unclear on this).
Can someone help me clean up the above thought process? It seems I didn't need to consider the characteristic polynomial at all. In general, how does knowledge of the minimal polynomial help you determine the dimension of a vector space?