Multiple eigenvectors for an eigenvalue and how to know We have the following matrix:
$$\begin{pmatrix}
    0 & -1 & 0 \\
    4 & 4 & 0 \\
    2 & 1 & 2
\end{pmatrix} $$
This matrix has characteristic equation $- \lambda ^3 + 6 \lambda ^2 - 12 \lambda + 8$, which gives us the eigenvalue $\lambda = 2$. However, this eigenvalue has two eigenvectors, namely $(1,-2,0)$ and $(0,0,1)$. 
How would you mathematically be able to know how many eigenvectors are associated with a single eigenvector? Is there a way to find out? 

The number of linearly independent eigenvectors that are associated with an eigenvalue, is called the geometric multiplicity of the eigenvalue. It can be found by solving the system $$\dim \ker(λΙ-Α)$$ 
In your case $λ=2$. To avoid confusion, the algebraic multiplicity of the eigenvalue is the number of times that this eigenvalue is a root of the characteristic polynomial. In your case $$p_A(λ)=-(λ-2)^3$$ so the algebraic multiplicity is equal to $3$. It is true that the geometric multiplicity is less or equal to the algebraic multiplicity.
