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

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?

• If the polynomial factors as $(x - \lambda)q(x)$ for some monic irreducible quadratic $q$, try looking at the null space of $q(T)$. Commented Nov 20, 2020 at 22:18
• I tried doing this, it would work out fine if $q(T)$ is not injective. If it is, we cannot pick a non trivial vector from the kernel. Can we ensure it will not be injective? Commented Nov 21, 2020 at 3:43

## 1 Answer

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.

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.