Getting a basis of eigenvalues $$\begin{bmatrix} a\\\\b\end{bmatrix} \longmapsto \begin{bmatrix}0&1\\\\-4&-4\end{bmatrix}\begin{bmatrix} a\\\\b\end{bmatrix}$$
Use the characteristic polynomail to find all eigenvalues for the transformation for each eigenvalues $\lambda$ find all eigenvectors with eigenvalues $\lambda$ and find a basis for $E_\lambda$
Here is what I have so far
$$\operatorname{char}(f)\begin{bmatrix}-x&1\\\\-4&-4-x\end{bmatrix}=x^2+4x+4 = (x+2)^2$$ so there is one eigenvalue $\lambda=-2$
I'm not sure how to find a basis for $E_{-2}$ though so this is as far as I have got.
 A: I think, you mistook a sign, it is rather $x^2+4x+4$.
You have to solve the equation
$$ \pmatrix{0&1\\-4&-4}\pmatrix{a\\b}=\pmatrix{-2a\\-2b}.$$ 
It gives $b=-2a$ and the second row: $-4a-4b=-2b$, that is, again, $-2a=b$.
So, the vectors of the form $\pmatrix{a\\-2a}$ are the eigenvectors. They span $1$ dimension, and a basis vector is $\pmatrix{1\\-2}$.
A: You just put the $\lambda$ back into the matrix (for x in your case) - to get the eigenvector corresponding to $\lambda = -2$. You get:
$$\begin{pmatrix}
 2 & 1 \\ 
 -4 & -2
 \end{pmatrix} $$  
But beware the eigenbasis of eigenvectors exists if and only if the algebraic multiplicity $\nu_a(\lambda)$ (that is the multiplicity of $\lambda$ as a root of the characteristic polynomial) is equal to the geometric multiplicity $\nu_g(\lambda)$ (which is the dimension of the null space of the above matrix). 
In this case $\nu_a(\lambda) = 2$. Using equivalent row operation we get matrix:
$$\begin{pmatrix}
 2 & 1 \\ 
 0 & 0
 \end{pmatrix} $$  
thus it's rank is 1 so the $\nu_g(\lambda) = 2-1 = 1 \ne \nu_a$ so the eigenbasis doesn't exist! 
Otherwise the solutions of the system with the matrix above (the solution are actually the eigenvectors) would form the eigenbasis, moreover the matrix is diagonalizable.
