Linear map $T: \mathbb{R^3} \to \mathbb{R^3}$ has a two dimensional invariant subspace

Question

I am supposed to show $T: \mathbb{R^3} \to \mathbb{R^3}$ has both a 1 dimensional and 2 dimensional invariant subspace, using the simplest machinery possible.

My approach: I have considered the 3-degree characteristic polynomial of $T$. If it has 3 real roots, we are done. If it has only one real root, that will be an eigenvalue and hence it's eigenspace is invariant 1D subspace. The natural choice for the 2D invariant subspace seems to be the orthogonal complement of that subspace, but I am unable to show that the image of that subspace under $T$ is still orthogonal to our 1D invariant subspace. Am I going in the right direction?

Answer 1

Let $\lambda$ be the real eigenvalue, and consider the subspace $W=\text{Range}(T - \lambda I)$.

Showing it is invariant: if $w \in W$, then it is of the form $w=(T-\lambda I)u$ for some $u \in \mathbb{R}^3$. Then $Tw=T(T-\lambda I) u = (T-\lambda I) T u \in W$.

Showing it has dimension $2$: use the rank-nullity theorem.

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Context: this subspace appears in the inductive proof of the Jordan form theorem. The real Jordan form of your matrix (in the case where the characteristic polynomial has one real root) will have two blocks of size 1 and 2 respectively, where the block of size two is the "real Jordan block" associated with the complex conjugate pair of complex eigenavlues.