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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.

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    $\begingroup$ If I'm calculating correctly, there's only one. $\endgroup$ – Matt Samuel Apr 15 at 18:26
  • $\begingroup$ I thought that there are at least 2... $\endgroup$ – user653342 Apr 15 at 18:31
  • $\begingroup$ You're right, there are at least $2$. $\endgroup$ – Matt Samuel Apr 15 at 18:48
  • $\begingroup$ 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 $\endgroup$ – user653342 Apr 15 at 18:51
  • $\begingroup$ 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. $\endgroup$ – Matt Samuel Apr 15 at 18:53
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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 $2$.

I'm guessing these are the two subspaces you found.

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