eigenvectors of a matrix with known eigenvalues $$
        \begin{matrix}
        1 & 1 & 0 & 0 \\
        0&1&0&0\\
        0&0&0&2\\
        0&0&2&0
        \end{matrix}
    $$
This matrix has eigenvalues 1, 2, & -2, I've solved the eigenvectors for 1 & 2, but not for -2. Subtracting (-2) from the diagonal produces:
$$
        \begin{matrix}
        3 & 1 & 0 & 0 \\
        0&3&0&0\\
        0&0&2&2\\
        0&0&2&2
        \end{matrix}
    $$
Solving the augmented matrix (not shown) gives me
$$
        \begin{matrix}
        0 \\
        0\\
        -1\\
        1\\
        \end{matrix}
    $$
but the answer is
   $$
        \begin{matrix}
        0 \\
        0\\
        1\\
        -1\\
        \end{matrix}
    $$
I've checked multiple sources and I'm wrong but I don't understand why. 
 A: As Amzoti said, "eigenvectors are not unique".
If you try to solve the following eigenproblem for $\textbf{x}=(x,y,z,t)$:
$$M\bf{x}=-2\bf{x}$$
where $M$ is your matrix, $-2$ is one of the eigenvalues and $\bf{x}$ is the eigenvector you're having trouble with, you'll find that $$z=-t$$
Therefore, the two eigenvectors you mention are equivalent, and a general expression for the eigenvector corresponding to the eigenvalue $-2$ is:
$$\mathbf{x}=(0,0,k,-k),\quad k\in\mathbb{R}^*$$
A: you can take advantage of the block structure of the matrix. you have two $2 \times 2$ matrices $B$ and $C$ where $B = \pmatrix{1&1\\0&1}, C = \pmatrix{0&2\\2&0}.$  $B$ has eigenvalues $1,1$  but only one eigenvector $(1,0)^T$ and $C$ has eigenvalues $-2, 2.$ the eigenvector corresponding to $-2$ is $(1,-1)^T$ and an eigenvector corresponding to $2$ is $(1, 1)^T.$ 
the eigenvalues of the mattrix $\pmatrix{B&0\\0&C}$ are $1,1, -2, 2.$ an eigenvector corresponding to the eigenvalue $1$ is $\pmatrix{1&0&0&0}^T,$  an eigenvector corresponding to the eigenvalue $-2$ is $\pmatrix{0&0&1&-1}^T$ and an
eigenvector corresponding to the eigenvalue $2$ is $\pmatrix{0&0&1&1}^T$
