Basis for the eigenspace of each eigenvalue, and eigenvectors So we want to find the basis for the eigenspace of each eigenvalue $\lambda$ for some matrix $A$. 
Through making this question, I have noticed that the basis for the eigenspace of a certain eigenvalue has some sort of connection to the eigenvector of said eigenvalue. Now I'm not sure if they actually equal each other, because I have some trouble when it comes to eigenvalues with a geometric multiplicity of two or more. 
Take the following example: 
$$\begin{pmatrix}
    0 & -1 & 0 \\
    4 & 4 & 0 \\
    2 & 1 & 2
\end{pmatrix} $$ 
This matrix has a characteristic polynomial $- \lambda ^3 + 6 \lambda ^2 - 12 \lambda + 8$. The root of this is $\lambda = 2$, which has an algebraic multiplicity of 3.
When I try to find the basis for the eigenspace of the eigenvalue $\lambda = 2$, I kind of get confused. Because when I solve $(A - 2I)\mathbf{v}$ I simply get $(0,0,1)$ (actually $(0,0,n)$ where $n \in \mathbb{R}$) as the answer for the basis, even though this eigenvalue has two associated linearly independent eigenvectors, namely $(0,0,1)$  and $(1,-2,0)$. This leaves me with the following questions:


*

*Is it true that the "basis of the eigenspace of the eigenvalue" is simply all of the eigenvectors of a certain eigenvalue (so in our example, the basis would be $(0,0,1), (1,-2,0)$)?

*If so, why am I not able to get both eigenvectors with my method? And how would I be able to get them both?   
 A: Your first question is correct, the "basis of the eigenspace of the eigenvalue" is simply all of the eigenvectors of a certain eigenvalue.
Something went wrong in calculating the basis for the eigenspace belonging to $\lambda=2$. To calculate eigenvectors, I usually inspect $(A-\lambda I)\textbf{v}=0$. In this case take $\textbf{v}=\left(\begin{array}{c}v_1\\v_2\\v_3\end{array}\right)$ to see:
$$(A-\lambda I)\textbf{v}=\left(\begin{array}{ccc}-2 & -1 & 0\\4 & 2 & 0\\2 & 1 & 0\end{array}\right)\left(\begin{array}{c}v_1\\v_2\\v_3\end{array}\right)=\left(\begin{array}{c}-2v_1-v_2\\4v_1+2v_2\\2v_1+v_2\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$
Note that all three equations are in fact the same (i.e. $A-\lambda I$ has three linearly dependent rows), such that the only restriction you have on $v_1,$ $v_2$ and $v_3$ is
$$2v_1=-v_2$$
Three unknowns with only one restriction leaves 2 degrees of freedom, i.e. you arrive at two eigenvectors. Now it is easy to see
$$\left(\begin{array}{c}1\\-2\\0\end{array}\right) \ \mbox{ and } \ \left(\begin{array}{c}0\\0\\1\end{array}\right)$$ 
are two of such eigenvectors.
