Does a repeated eigenvalue always mean that there is an eigenplane under the transformation matrix? If you have a 3x3 matrix, if you find that it has repeated eigenvalues, does this mean that there is an invariant plane (or plane of invariant points if eigenvalue=1)?
I always thought that there was an invariant plane if all 3 equations were the same when trying to find the eigenvectors, but does this only happen when there is a repeated eigenvalue, or does it happen also when there are 3 distinct eigenvalues? 

If $Av=\lambda v$ and $Aw=\lambda w$, then for any linear combination $\alpha v+\beta w$ we have 
$$
A(\alpha v+\beta w)=\alpha Av+\beta Aw=\alpha\lambda v+\beta\lambda w=\lambda(\alpha v+\beta w). 
$$
In words, a linear combination of eigenvectors for the same eigenvalue is again an eigenvector for that eigenvalue. 
That said, it could happen that no such linearly independent $v$ and $w$ exist: let 
$$
A=\begin{bmatrix} 2&1&0\\0&2&0\\0&0&3\end{bmatrix}.
$$
Then, while $2$ is a repeated eigenvalue, its eigenspace is one-dimensional. 
