Linear Algebra - Invariant Subspaces in 2-dimensions

Question

How do I find 2-dimensional subspaces that are invariant subspace of T ?

$T\left(\left[ \begin{matrix} x\\ y\\ z\\ \end{matrix} \right]\right) = \left[\begin{matrix} 2y+z \\ -2x+4y+z \\ -2x+2y+3z \end{matrix}\right]$

Answer 1

Here's a Axler technic:

Choose $v=(0,0,1) \in \Bbb R^3$. Then the set $\{v,Tv,T^2v,T^3v\}$ is dependent. Actually the set is $$\{(0,0,1),(1,1,3),(5,5,9),(19,19,27)\}$$ Here $$-(0,0,1)+\frac{5}{6}(1,1,3)-\frac{1}{6}(5,5,9)+0(19,19,27)=0$$ That is $$(-1)v+\frac{5}{6}Tv-\frac{1}{6}T^2v+0 T^3 v=0$$ That is $$\left( -\frac{1}{6} T^2+\frac{5}{6}T-I \right) v=0$$ Now it is easy to chech that $\text{span} \{v,Tv\}$ is a invariant subspace of dimension two!

Answer 2

here's some hint.

I think you can easily find the matrix expression of the given map.

and then try to find eigenvectors. since eigenspaces are T-invariant, you'd find 2-dimensional subspace.