Compute the eigenvalues and eigenvectors of A. 
Are there any convenient ways to compute the eigenvalues and eigenvectors of this matrix?
 A: One convenient way to compute the eigenvalues and eigenvectors of this matrix is to use the fact that it is a circulant matrix. An $n \times n$ circulant matrix is any matrix of the form  

You can show that the eigenvalues and eigenvectors are of the form
\begin{align}
v_j &= \frac{1}{\sqrt{n}} \left( 1, \omega_j, \cdots \omega_j^{n-1} \right)^T\\
\lambda_j &= \displaystyle \Sigma_{i=0}^{n-1} c_i \omega_j^{n-i} \\
\end{align}
for $j=0, \cdots n-1$ where $\omega_j = \exp \left(\frac{2 \pi i j}{n}\right)$. So using this fact, we can easily compute that the two eigenvalues of $A$ are $-2$ with multiplicity $n-1$ and $2(n-1)$ with multiplicity $1$. Again, using the properties of circulant matrices we get that the eigenspace corresponding to $\lambda = 2(n-1)$ is $span( \mathbb{1})$ where $\mathbb{1}$ is the all-ones vector. And so the eignenspace for $\lambda = -2$ is the set of vectors in $\mathbb{R}^n$ orthogonal to $\mathbb{1}$.
A: First of all, observe that $A+2I$ has $n$ columns of all $2$s, which will make $\mathrm{dim ~ Null}(A+2I) = n-1$, showing that $-2$ is an eigenvalue as long as $n > 1$. (If $n = 1$, $A$ is the zero transformation which has only eigenvalue $0$).  
We find the remaining eigenvalue next:
Consider the vector $ v = \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}$
Then matrix multiplication will show that: $Av = \begin{bmatrix} 2(n-1) \\ 2(n-1) \\ \vdots \\ 2(n-1) \end{bmatrix}$.
Observe that $Av = 2(n-1)v$, making $v$ an eigenvector with eigenvalue $2(n-1)$.
Since $\mathrm{dim} ~ E(-2, A) = n-1$, $\mathrm{dim} ~ E(2(n-1), A) = 1$,
We conclude that $A$ is diagonalizable and $E(2(n-1), A) = \mathrm{span}(v)$
We need only compute the eigenvectors associated with $-2$
For the purpose of (what I think is) ease of formatting, let $e_1, ... , e_n$ be the basis $A$ is written with respect to. Consider a vector of the form $w_k = e_1 - e_k$ for $k \in \{2, ..., n\}$. Then $A(e_1 - e_k) = 2e_2 + ... + 2e_n - (2e_1 + ... + 2e_{k-1} + 2e_{k+1} + ... + 2e_n) = -2e_1 + 2e_k$. 
So each $w_k$ is an eigenvector with eigenvalue $-2$ 
(and it is easy to see linear independence of the $w_k$'s).
Thus $E(-2, A) = \mathrm{span}(e_1 - e_2, e_1 - e_3, ... , e_1 - e_n)$, or, in alternate form:
$$\mathrm{span}\left(\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \\ \vdots  \\ 0 \end{bmatrix} , ... , \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ -1 \end{bmatrix} \right)$$
Note: I came up with this solution by clever reasoning of which vectors would be eigenvectors. I wanted to include my answer as a possible method to the problem (and it does answer the question), but there is probably a more systematic method to approach this problem than these tricks.
A: Christian found all but one of the eigenvalues by showing that $A + 2I$ had nullity $n-1$ ($v_i = e_1 - e_i$ for $2 \leq i \leq n$ span the $2$-eigenspace.)
Also, when guessing in general for eigenvalues there are a few tricks that one gets used to when the matrix looks nice. One is looking for how to make a singular matrix by adding a multiple of the identity (as Christian did). Another is saying that all the entries of each row add to $2(n-1)$ so $v = [1 \cdots 1]^T$ is an eigenvector with eigenvalue $2n - 2$. If instead all the columns added to a fixed constant then that common value would also be a constant since a matrix is similar to its transpose.
So, the eigenvalues are $2n - 2$ and $-2$ with respective eigenvectors $v$ and $v_i$, $2 \leq i \leq n$.
