This answer is confusing $4\times 4$ eigenvalue calculation Question:

Find the rank and the four eigenvalues of the following matrix:
$\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1  \end{bmatrix}$

Answer:

The matrix has rank 2 (ensuring two zero eigenvalues) and $(1,1,1,1)$ is an
  eigenvector with  $\lambda=2$. With trace 4, the other eigenvalue is
  also $\lambda=2$, and it's eigenvector is $(1,-1,1,-1)$.

This is directly taken from my book. The answer does not give a clear answer as to what the four eigenvalues are.


*

*Is the answer $0,0,2,2$(the four eigenvalues)?

*I see that the rank is $2$ and the trace is $4$. $n-r=2$, meaning that two of the eigenvalues are $0$. The determinant is $0$. I know that the trace is the sum of the eigenvalues, so the sum in this case is 4. How does one calculate the two eigenvalues?

 A: The answer in your book says "two zero eigenvalues ... an eigenvector with $\lambda = 2$ ... the other eigenvalue is also $\lambda = 2$". 
So it says the eigenvalues are $0,0,2,2$. How is this not clear?
As for how to compute them (without computing a 4x4 characteristic polynomial), you're almost there. The rank shows that two of the eigenvalues are $0$, so you have two more to find, call them $\lambda_1$ and $\lambda_2$. Now the trace is $4$, so $0+0+\lambda_1+\lambda_2 = 4$. So if you can find $\lambda_1$, you'll have $\lambda_2 = 4-\lambda_1$.
Now you observe that $(1,1,1,1)$ is an eigenvector (How would you observe that? Well, it's true for any matrix all of whose rows sum to the same number). It has eigenvalue $\lambda_1 = 2$, so $\lambda_2 = 2$ also.
A: If I compute $\text{det}(M-I\lambda)$, where $M$ is the matrix, $I$ the identity matrix and $\lambda$ and eigenvalue, I get:
$$\text{det}(M-I\lambda)=\begin{vmatrix}1-\lambda & 0 & 1 & 0\\ 0 & 1-\lambda & 0 & 1\\ 1 & 0 & 1-\lambda & 0\\ 0 & 1 & 0 & 1-\lambda\end{vmatrix}=4\lambda^2-4\lambda^3+\lambda^4$$
The eigenvalues must satisfy $\text{det}(M-I\lambda)=0$. Since this factors to $(\lambda-2)^2\lambda^2$, there are two eigenvalues of 2 and two eigenvalues of zero.
