Try to solve without Jordan theory. $f \in End(\mathbb{R^3})$ prove that $\exists W, \; dim(W)=2$ invariant subspace.

Question

Using the Jordan normal form for real spaces it should be easy, because the first two vectors of every basis that gives a Jordan matrix generate a 2-dimensional invariant subspace.

Any idea about how to solve it without Jordan?

Thanks in advance :)

Answer 1

The characteristic polynomial $\pi$ of $f$ is of $\deg$ 3, so it admits a real root. We can write $\pi = (x -\lambda)^{n}Q$, with $\gcd(x-\lambda, Q) = 1$ and $\deg(Q) = 0, 1,\text{ or } 2$.

• if $n = 3$. Then $\dim(\ker(f - \lambda Id)^3) = 3$, and $\dim(\ker(f - \lambda Id)) \ge 1$. Take $W = \ker(f - \lambda Id)$ or $W =\ker(f - \lambda Id)^2$
• if $n < 3$ so $\deg(Q) \ge 1$, we have $\dim(\ker(f -\lambda Id)^n)\ge 1$, $\dim(\ker(Q(f)) \ge 1$ and $\dim(\ker(f -\lambda Id)^n) + \dim(\ker(Q(f)) = 3$. So $\dim(\ker(Q(f)) = 2$ or $\dim(\ker(f -\lambda Id)^n) = 2$. take $W$ to be $\ker(Q(f))$ or $\ker(f -\lambda Id)^n$ accordingly.