Suppose we have a linear map $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ whose minimal polynomial is $m_T(x)= x^2+1$. We cannot apply the primary decomposition theorem to give non-trivial T-invariant subspaces, since the roots are imaginary.
Does this allow us to conclude that there are no non-trivial T-invariant subspaces of $\mathbb{R}^2$ or does it only mean that we cannot apply the primary decomposition theorem and that non-trivial T-invariant subspaces can still be found by a more direct approach (i.e. by using matrix representation of T to find which lines are invariant)?
Many thanks.
Assume that $U$ is non-trivial T-invariant subspace. Then $\dim U =1$, hence there is $x_0 \ne 0$ such that $U = span \{x_0\}.$
From $T(x_0) \in U$ we get some $ \lambda \in \mathbb R$ with $T(x_0)= \lambda x_0.$
But then $ \lambda $ is an eigenvalue of $T$, hence $\lambda = \pm i,$ a contradiction.