Let $V$ be a finite-dimensional vector space over $\mathbb{R}$, and let $T: V \rightarrow V$ be a linear transformation such that $T^{4} - I = T^{3} + 3T^{2} + T + 3 =0$. Prove that $V$ is a direct sum of $2$-dimensional $T$-invariant subspaces.
$T$ satisfies: $f(x) = (x-1)(x+1)(x^{2} + 1)$ and $g(x) = (x + 3)(x^{2} + 1)$. Hence, the minimal polynomial of $T$ must divide both $f$ and $g$ and it follows that $m_{T}(x) = x^{2} + 1$.
I know that $V = \mathbb{R}^{\dim V} = \bigoplus_{i = 1}^{r}\mathbb{R}[x]/(p_{i}(x))$, where $p_{i}|p_{i+1}$ and the largest of these is $m_{T}(x)$ and since $m_{T}(x)$ is irreducible over $\mathbb{R}$, $p_{i} = x^{2} + 1$ for all $i$.
If I can conclude that the dimension of $V$ is $4$, then I am done as then there will be only another invariant factor, which must be also equal to the minimal polynomial.
Another way one could proceed is to invoke the primary decomposition to write $\ker (T + 3I) \oplus \ker (T^{2} + I) = V = \ker (T-I) \oplus \ker (T+I) \oplus \ker (T^{2} + I)$.