$\vec v_0,...,\vec v_{n-1}$ is a basis in a linear space. $A\vec v_i=\vec v_{(i-1)~mod~n}+\vec v_{(i+1)~mod~n}$. Find eigenvalues of $A$. Suppose $\vec v_0,...,\vec v_{n-1}$ is a basis in linear space. $A$ - is a linear operator such that $A\vec v_i=\vec v_{(i-1)~mod~n}+\vec v_{(i+1)~mod~n}$. Find eigenvalues of $A$.

I assuming that in a case of $i=0$ that means: $A\vec v_0=\vec v_{n-1} + \vec v_1$.
In a case of n=3:
$$
    \begin{cases}
      A\vec v_0=\vec v_2 + \vec v_1\\
      A\vec v_1=\vec v_0 + \vec v_2\\
      A\vec v_2=\vec v_1 + \vec v_0\\
    \end{cases}
$$
We can notice that each $\vec v_i$ presented in exactly $2$ equations, so we can write:
$A\times(\vec v_0 + \vec v_1 + \vec v_2) =\vec v_2 + \vec v_1+\vec v_0 + \vec v_2+\vec v_1 + \vec v_0= 2 \times(\vec v_0 + \vec v_1 + \vec v_2)$
so the $2$ is an eigenvalue.
Similarly, we can notice that:
$A\times (2\vec v_0-\vec v_1 - \vec v_2)=2\vec v_2+2\vec v_1 -\vec v_0 -\vec v_2-\vec v_1 - \vec v_0=-1\times (2\vec v_0-\vec v_1 - \vec v_2)$
So the $-1$ is an eigenvalue.
I'm confused that we can make the last trick with $(-\vec v_0+2\vec v_1 - \vec v_2)$ and $(-\vec v_0-\vec v_1+2\vec v_2)$ getting the same $-1$ eigenvalue.
So in total we have $4$ eigenvectors for $n\times n$ matrix?
 A: The eigenvalues are $2$ and $\lambda_j=2\cos 2\pi j/n$ for $j=1,\ldots,\lfloor n/2 \rfloor$. Look for corresponding eigenvectors of the form $ \varphi_j(k)=\cos(2\pi jk/n)$
and $ \psi_j(k)=\sin(2\pi jk/n)$.
When $n$ is odd, each $\lambda_j$ has multiplicity 2.
When $n$ is even, only $\varphi_j$ appears for $j=n/2$.
A: You have found the correct eigenvalues, and the eigenvectors you constructed for $-1$ are correct. The catch is that the three vectors are not linearly independent. If you add the three vectors $(2\vec v_0-\vec v_1 - \vec v_2)$, $(-\vec v_0+2\vec v_1 - \vec v_2)$, and $(-\vec v_0-\vec v_1+2\vec v_2)$ together, you get zero. You are right that the number of eigenvalues and eigenvectors of a matrix should be bounded by the size of the matrix, but the precise statement is that the total geometric multiplicity (sum of the maximum number of linearly independent eigenvectors for each eigenvalue) is at most the dimension of the matrix. In this case, the geometric multiplicity of the eigenvalue $-1$ is $2$, because while the three eigenvectors you found are not linearly independent, removing any one of them leaves the remaining set linearly independent.
