Let $\alpha$ be a real number, find all invariant subspaces for the matrix $$ \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ How does the result depend on $\alpha$?

I am a bit confused about how to find all the spaces, I can see that if $\alpha=0$ then every subspace is invariant, but what do I do in other cases? Should I find the eigenspaces?

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    $\begingroup$ Picture the transformation: it is a rotation about the z-axis. $\endgroup$ – Trevor Gunn Nov 13 '17 at 18:47
  • $\begingroup$ Notice this matrix is always inevitable and always has determinant 1. $\endgroup$ – Wintermute Nov 13 '17 at 18:50
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    $\begingroup$ An "invariant subspace" of a matrix is a subspace such that any vector, v, in that subspace has the property Av is also in that same subspace. This family of matrices is the set of all rotations around the z-axis. The invariant space is the z-axis, (0, 0, z). I shall now sit and meditate on "inevitable matrixes"! $\endgroup$ – user247327 Nov 13 '17 at 18:53
  • $\begingroup$ @user247327 I have no idea what you mean by "inevitable matrixes" $\endgroup$ – Omnomnomnom Nov 13 '17 at 18:54
  • $\begingroup$ @Omnomnomnom: Pretty sure it was a typo and meant to say "invertible". $\endgroup$ – Michael Joyce Nov 14 '17 at 22:45

We assume that $\alpha\notin \pi\mathbb{Z}$. Your matrix, say $R$, has $3$ distinct eigenvalues over $\mathbb{C}$: $e^{i\alpha},e^{-i\alpha},1$ with associated eigenvectors $v,\bar{v},e_3$. A general result says that the proper invariant subspaces over $\mathbb{C}$ are the $span(U)$, where $U$ goes through the strict subsets of $\{v,\bar{v},e_3\}$ (there are $6$ such vector spaces).

Over $\mathbb{R}$, you must group the two first eigenvectors and you cannot group $v,e_3$ or $\bar{v},e_3$. Then, there are only two proper invariant spaces: $span_{\mathbb{R}}(e_1,e_2)\subset span_{\mathbb{C}}(v,\bar{v})$ and $span(e_3)$.


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