How to find the eigenvalue? I have been given a matrix 
$$A = \begin{bmatrix}
1&1&0&0\\
1&1&0&0\\
0&0&1&1\\
0&0&1&1
\end{bmatrix}$$
I expanded by row one twice to get the characteristic polynomial: 
$(1-\lambda)^2[(1-\lambda)^2 -1] - 1[(1-\lambda)^2 - 1]$
Which I solved lambda and got that $\lambda = 0$. I checked my work with the answer in the back and its wrong. They got another value for lambda which I can't seem to get. 
http://i1317.photobucket.com/albums/t638/ayoshnav/Snapshot_20140326_zps80f4ba36.jpg
http://i1317.photobucket.com/albums/t638/ayoshnav/Snapshot_20140326_1_zpsf7f28df4.jpg

Another approach would be to explicitly find eigenvectors - given the form of your matrix, it's quite easy.
First, the vectors $$\begin{pmatrix}1\\1\\0\\0\end{pmatrix},\quad \begin{pmatrix}0\\0\\1\\1\end{pmatrix}$$
are obviously eigenvectors; the corresponding eigenvalue is $2$.
Then again,
the vectors $$\begin{pmatrix}1\\-1\\0\\0\end{pmatrix},\quad \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$$ are eigenvectors, too, with eigenvalue $0$.
We've found $4$ different eigenvectors, the matrix is $4\times 4$, hence we can conclude that we found all eigevalues.
