Find the eigenvalues of A and a basis for each eigenspace of A. 
Let $A$ =
  $$
        \begin{bmatrix}
        -4 & -4 & 0 \\
        2 & 2 & 0 \\
        2 & 2 & 0 \\
        \end{bmatrix}
$$
  Find the eigenvalues and eigenspaces of $A$

I've got the eigenvalues to be $0$ and $-2$, and I have got the eigenspaces corresponding to the eigenvlues to be
For $0$ = $$
        \begin{bmatrix}
        -1  \\
        1  \\
        0  \\
        \end{bmatrix}
$$
For $-2$ = $$
        \begin{bmatrix}
        -2  \\
        1  \\
        1  \\
        \end{bmatrix}
$$
However the solution says the eigenspace for $0$ is $$
        \begin{bmatrix}
        0  \\
        0 \\
        1  \\
        \end{bmatrix}
$$ and $$
        \begin{bmatrix}
        -1  \\
        1  \\
        0  \\
        \end{bmatrix}
$$
Why is it that?
 A: Recall that the columns of a transformation matrix are the images of the basis vectors. The last column of $A$ tells us that the image of $(0,0,1)^T$ is zero, so it is an eigenvalue of $0$.  
The result of right-multiplying a matrix by a column vector is a linear combination of its columns. We can see that subtracting the second column from the first gives zero, so the image of $(1,-1,0)$ is zero, which means that it, too is an eigenvector of $0$. This vector is not a multiple of $(0,0,1)^T$, so we know that $0$ has both algebraic and geometric multiplicities of at least two, and that these vectors can form part of a basis for its eigenspace.  
The sum of the eigenvalues, taking into account their multiplicities, is equal to the trace of the matrix. Here, $\operatorname{tr}A=-4+2+0=-2=-2+0+0$, which gives us the third eigenvalue. I’ll leave finding a corresponding eigenvector to you.
A: The eigenspace relative to $0$ can be deduced from the RREF of the matrix, which is
$$
\begin{bmatrix}
1 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
This shows there are two free variables; the only equation is $x_1+x_2=0$, so a basis of the eigenspace is obtained by first choosing $x_2=1$ and $x_3=0$, then $x_2=0$ and $x_3=1$:
$$
\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix},
\qquad
\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
$$
