# all 2 dimensional invariant subspaces

How we can find all 2 dimensional invariant subspaces of $$\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \end{pmatrix}$$

I know that there are at least 2 such subspaces, but I don't know if there are exactly 2 (and if it is true how to prove it) or if there are more than 2.

• If I'm calculating correctly, there's only one. – Matt Samuel Apr 15 at 18:26
• I thought that there are at least 2... – user653342 Apr 15 at 18:31
• You're right, there are at least $2$. – Matt Samuel Apr 15 at 18:48
• So I think that I know 2, but I don't know if there are any other ones or how to prove that there are exactly 2 such subspaces – user653342 Apr 15 at 18:51
• You should include that information in your question specifically. Spell out what you know so others only need to fill in what you don't. – Matt Samuel Apr 15 at 18:53

Any two invariant two dimensional subspaces must intersect in an eigenvector. If such a subspace contains $$e=(0\ 0\ 1)^t$$, suppose $$v$$ is a vector in the subspace that is not a multiple of $$e$$. The only possibility is for the second component to be $$0$$, otherwise $$v,Av, e$$ is a basis for the whole space. Thus there is only one invariant two dimensional subspace containing $$e$$, and it contains the other eigenvector. Call this subspace $$U$$.
If $$U'$$ is another invariant two dimensional subspace it follows that $$U\cap U'$$ is the one dimensional subspace spanned by $$(1\ 0\ 0)^t$$. Let $$v'=(x\ y\ z) ^t$$ be a vector completing a basis for $$U'$$. Then necessarily $$y\neq 0$$. If $$z\neq 0$$, then $$U'$$ contains $$e$$ as well, which is impossible. Thus the second invariant subspace, the only other one, is the one corresponding to the Jordan block with eigenvalue $$2$$.