Finding eigenvectors of matrix of matrices 
Let $A$ be a matrix with $\lambda_1,...,\lambda_n$ eigenvalues, and $v_1,...,v_n$ the corresponding eigvenvectors. Let $B= \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix}$. It's known that the eigenvalues of $B$ are $\{\pm \lambda_1,..., \pm \lambda_n\}$. Find the eigenvectors of $B$.

I've managed to find $n$ of the $2n$ eigenvectors of $B$, those who correspond to the eigenvalues: $\{ \lambda_1,...,  \lambda_n\}$:
$u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$. I assumed that the new eigenvectors should be built of the original eigenvectors, thus I started playing with combinations of the original eigenvectors, such as: $\begin{pmatrix} v_i \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ v_i \end{pmatrix}$ however those vectors didn't help at all. Eventually, looking at how $B$ is defined, I understood that since $B$ simply works the same as $A$ works on vectors for the first half and then again for the second half, I found that  $u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$ can work.
However, I couldn't  manage to find the corresponding eigenvectors for the $\{-\lambda_1,...,-\lambda_n\}$ eigenvalues. I can't see how applying the original matrix $A$ in any kind of form will get us to the negative of its' eigenvalues.
Any hint/clue?

Let's look at $u_i =\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$:
$B \begin{pmatrix} v_i \\ -v_i \end{pmatrix} = \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix} \begin{pmatrix} v_i \\ -v_i \end{pmatrix}= \begin{pmatrix} A(-v_i) \\ Av_i \end{pmatrix} = \begin{pmatrix} -\lambda_iv_i \\ \lambda_iv_i \end{pmatrix} = -\lambda_i\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$
Thus we get the other $n$ eigenvectors, ending with total of $2n$ eigenvectors.
