Find the Eigen values and Eigen vectors of a 4x4 symmetric matrix The matrix is $$
\begin{bmatrix} 
2 & 0 & -2 & 0 \\
0 & 2 & 0 & -2 \\
-2 & 0 & 2 & 0 \\
0 & -2 & 0 & 2 \\
\end{bmatrix}$$
To get the characteristic polynomial, I used $\sum_{r=0}^4\text{tr}_{n-r}(A)(-t)^r$. But this process is lengthy. Is there something I am overlooking? 
 A: Observe the matrix is of the block form $$2\begin{pmatrix}I&-I\\-I&I\end{pmatrix}. $$
So we can look at the matrix $$2\begin{pmatrix}1&-1\\-1&1\end{pmatrix} $$ first. This is much easier to do either by hand calculation or by inspection, and we get that the eigenvalues are $0$ and $4$ with eigenvectors $(1,1)^T$ and $(1,-1)^T.$ 
So we need to figure how these eigenvectors translate into eigenvectors of the four dimensional matrix.  Since all the blocks are proportional to the identity, we can just change the $1$'s in the above 2D eigenvectors to any vector and get a valid eigenvector of the 4D matrix. So $(a,b,a,b)^T$ is an eigenvector with eigenvalue $0$ an and $(a,b,-a,-b)^T$ is an eigenvector with eigenvalue $4.$ 
Since each of the eigenvalues has a two-dimensional space of corresponding eigenvectors we have found all the eigenvectors, and the eigenvalues $0$ and $4$ each have multiplicity 2. The eigenvectors can be chosen as any linearly independent options for $a$ and $b$ above. So, for instance, we can be boring and take $(a,b) = (1,0)$ and $(0,1)$ and get eigenvectors $(1,0,1,0)^T$ and $(0,1,0,1)^T$ corresponding to the eigenvalue $0$, and then $(1,0,-1,0)^T$ and $(0,1,0,-1)$ corresponding to eigenvalue $4.$ 
Note that there are many possible choices for the eigenvectors. You can choose any $(a_1,b_1)$ and $(a_2,b_2)$ that are linearly independent.
A: The eigen values are 0,0,4,4 and corresponding linearly independent eigen vectors are (1,0,1,0),(0,1,0,1),(1,0,-1,0),(0,1,0,-1).
A: Note that for $B= \frac{1}{4}A$, it holds $B^2 = B$. Hence, if $\lambda$ is an eigenvalue of $A$, then $$\Big(\frac{\lambda}{4}\Big)^2 = \frac{\lambda}{4}\qquad \implies \qquad \lambda^2 = 4\lambda \qquad \implies \qquad \lambda\in\{0,4\}.$$ Now, you need to solve the systems $Av=0$ and $Aw-4w=0$ for $v,w$ to find the corresponding eigenvectors.
