Existence of invariant plane for repeated eigenvalues If we are in $\mathbb{R^3}$ and the characteristic equation has a repeated root, does that always mean that there is an invariant plane rather than just an invariant line, like when there is no repeated root?
 A: No. The dimension of the eigenspace associated with $\lambda$ is called the geometric multiplicity of $\lambda$. It is a well-known fact that geometric multiplicity is less than the algebraic multiplicity, where algebraic multiplicity is referring to the number of repetitions of $\lambda$ in the characteristic equation.
A: Suppose $A$ is a $3\times3$ matrix. $A$ must have at least one real eigenvalue. There are only $2$ cases.

$A$ has at least $2$ linearly independent eigenvectors

Suppose $\vec x, \vec y$ are $2$ linearly independent eigenvectors. The plane $
\langle\vec x,\vec y\rangle=\{a\vec x+b\vec y:a,b\in\mathbb{R}\}$ must be invariant under $A$ i.e. $A(\langle\vec x,\vec y\rangle)=\langle\vec x,\vec y\rangle$.

$A$ has $1$ real eigenvalue and the dimension of the eigenspace is $1$

$A$ is similar to 
$$
\begin{bmatrix}\lambda&1&0\\0&\lambda&1\\0&0&\lambda\end{bmatrix}
$$
under which $\langle[1,0,0]^\top,[0,1,0]^\top\rangle$ or the plane $z=0$ is invariant.
So choosing the right basis you can always have an invariant plane.
