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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?

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    You can bring the matrix in block-diagonal form by applying a similarity transform (permutation ...) – 2017-02-21

3 Answers 3

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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.

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    To formalize these observations yet further: we can note that the matrix is $$ 2\pmatrix{1&-1\\-1&1} \otimes I $$ where $\otimes$ denotes the [Kronecker product](https://en.wikipedia.org/wiki/Kronecker_product). – 2017-02-21
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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).

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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.