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? 
 A: 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. 
A: For any $n \times n$ (complex) matrix $A$, there are always invariant subspaces of $\mathbb C^n$  of all dimensions $\le n$.  If $A$ is upper triangular, the $k$-dimensional subspace spanned by the first $k$ standard unit vectors is invariant.  And every square matrix is similar to an upper triangular matrix.
A: Note: I realized eigenplan is the space formed by the eigenvectors for a repeated eigenvalue . So the title you ask a bout  the existence of a   eigenplan, but in the text you speak  a bout invariant hyperplane . Then an invariant hyperplane is not necessarily a eigenplan, the converse is yes.
The answer to the question given in the text, is already indicated, even more in the answer you get better.
But namely when a eigenvalu is multiple of multiplicity m, the dimension of the  associated eigenspace to this eigenvalue  is bounded below by 1 and superiorly by m.
So all cases are possible, for example, look at the type matrix of 3x3 , of Jordan form matrix.
if my remark is false I apologize for my understanding of English, thanks
