Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ be the linear operator
$ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $
The problem I am trying to understand is the following.
True or False? If $W$ is a $T$-invariant subspace of $\mathbb{R}^3$ $\exists T$-invariant subspace W' of $\mathbb{R}^3$ such that W \oplus W' = \mathbb{R}^3. I think the answer is true and I will list my ideas below but I think there must be an easier way to do approach this.
Since $A$ has distinct eigenvalues $1,2,3$ we see the minimal polynomial is $m_A(x) = (x-1)(x-2)(x-3)$
and therefore using the fundamental structure theorem for PID's we have $\mathbb{R}^3 = \mathbb{R}/(x-1)\oplus\mathbb{R}/(x-2)\oplus \mathbb{R}/(x-3) = \mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R} $
From this calculation does it follow that W' = \mathbb{R}\oplus\mathbb{R} is the invariant subspace we are looking for?
Is there a way to do this problem without using the fundamental structure theorem for modules over a PID?