Determine the eigenspace Given the matrix $$A =\begin{pmatrix}-3&0&1&0\\0&-3&0&1\\-1-b&-4&-4&-4\\b&3&3&3\end{pmatrix}$$ 
I have to check whether $\lambda = -2$ is an eigenvalue of the matrix. If this is the case, I have to determine the eigenspace for all real b. 
So I put it in: $$ A - \lambda I = A - (-2)I = A + 2I $$ 
and used Gauss-elimination to get the row reduced form of the new matrix:
$$\begin{pmatrix}1&0&0&\frac{8}{b+3}\\0&1&0&-1\\0&0&1&\frac{8}{b+3}\\0&0&0&0\end{pmatrix}$$ 
Since $ rang(A + 2I) = 3 < 4 = rang(A) $, I know that it is indeed an eigenvalue, albeit I don't quite understand why that would be the case. 
How do I get the eigenspace from this point?
Maybe the first vector would be 
$$ \begin{pmatrix} -\frac{8}{b+3}\\1\\-\frac{8}{b+3}\\1\end{pmatrix} $$
But I wouldn't understand why, and how to get the other (eigen-?)vectors. Why do I put a one, where the row contains only zeros? How do I decide what the other vectors should contain to get an eigenspace?
 A: It helps to read what your row reduced echelon form actually says, in terms of equations. It says:
$$x_1+\frac{8}{b+3} x_4 = 0. \\
x_2 - x_4 = 0. \\
x_3 + \frac{8}{b+3} x_4 = 0. \\
0 = 0.$$
That last equation is trivial, and the other three equations are independent of one another (since only the first involves $x_1$, only the second involves $x_2$, and only the third involves $x_3$). Thus the solution to the problem is unique once you specify $x_4$. We say that $x_1,x_2,x_3$ are dependent variables and $x_4$ is an independent or free variable. Thus you uniquely specify an eigenvector once you decide on a nonzero value for $x_4$. Here they chose $1$, but it doesn't really matter what it is. You find the other entries by plugging in the value you chose into the system and solving it by back-substitution. 
Since you have just one free variable, all the eigenvectors in this eigenspace are multiples of this "representative" eigenvector that you have written down. In general you have as many linearly independent eigenvectors as you have free variables. You can find them by choosing linearly independent values of the free variables; one way is to make all but one free variable zero and one of them $1$, varying that special one among all the free variables.
A: You computed the RREF correctly, apart from the fact you need to distinguish the case when $b+3\ne0$ and when $b+3=0$.
For $b+3\ne0$, the rank of the matrix is $3$, so every nonzero vector in its null space is an eigenvector of $A$ relative to $-2$. Indeed, if $(A+2I)v=0$, we have $Av+2v=0$, so $Av=-2v$, as requested for $v$ to be in the eigenspace.
Since the null space has dimension $1$, just pick one nonzero vector in it: the system to solve is
$$
\begin{cases}
x_1=-\dfrac{8}{b+3}x_4\\[4px]
x_2=x_4\\[4px]
x_3=-\dfrac{8}{b+3}x_4
\end{cases}
$$
and you can choose as a basis for the eigenspace the vector you get by setting $x_4=b+3$, so
$$
\begin{bmatrix}-8\\b+3\\-8\\b+3\end{bmatrix}
$$
For $b+3=0$, the RREF is
$$
\begin{bmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
so the system to solve is
$$
\begin{cases}
x_1=x_3 \\[3px]
x_2=0 \\[3px]
x_4=0
\end{cases}
$$
and a basis for the eigenspace is obtained by setting $x_3=1$, that is, the vector
$$
\begin{bmatrix}1\\0\\1\\0\end{bmatrix}
$$
However, if we choose $x_3=-8$, we get the same vector as before, with $b=-3$, so in this particular case a basis for the eigenspace can be chosen as consisting of the vector
$$
\begin{bmatrix}-8\\b+3\\-8\\b+3\end{bmatrix}
$$
for all values of $b$.
