How do I prove that the inverse of the matrix, M, below has all elements greater than or equal to 0? $M=
  \begin{bmatrix}
    1 & \nu & 0 & ... & 0 & -\nu \\
    -\nu & 1 & \nu & 0 & ... & 0 \\
    0 & -\nu & 1 & \nu & ... & 0 \\
    0 & 0 & -\nu & 1 & \nu.. & 0 \\
     .&  &  .& .& .&   \\ 
    \nu & 0 & ... & 0 & -\nu & 1 \\
  \end{bmatrix}
$
It is basically a Circulant matrix.
How do I prove that the inverse of this matrix will have all positive entries. For what conditions of $\nu$ will it be a non-negative matrix ?
 A: (Needs too much space to be a comment, so an answer it is)
As in the case of all circulant matrices you can find the inverse by using the discrete Fourier transform. The eigenvectors of the $N\times N$ version $M(N,\nu)$ of your matrix are $(1,\zeta^j,\zeta^{2j},\ldots,\zeta^{(N_1)j})^T$, where $j=0,1,\ldots,N-1,$
$\zeta=e^{2\pi i/N}$ and
$$
M(N,\nu)x_j=\lambda_jx_j,\qquad \lambda_j=(1-2i\nu\sin\frac{2\pi j}N).
$$
The inverse $M(n,\nu)^{-1}$ is then the circulant matrix with coefficients gotten from the IDFT of $(1/\lambda_0,1/\lambda_1,\ldots,1/\lambda_{N-1})$.
But before we try to calculate that let's test some cases. Set $N=4$. Also sprach Mathematica:
$$
(1+4\nu^2)M(4,\nu)^{-1}=
\left(\begin{array}{cccc}
1+2\nu^2&-\nu&2\nu^2&\nu\\
\nu&1+2\nu^2&-\nu&2\nu^2\\
2\nu^2&\nu&1+2\nu^2&-\nu\\
-\nu&2\nu^2&\nu&1+2\nu^2
\end{array}\right)
$$
Both $\nu$ and $-\nu$ appear as coefficients, so I don't see a way of making all the entries positive here.
Please check the question.
A: A quick observation: when the size $N$ of the matrix is even, $M^{-1}$ can never be entrywise positive.
As pointed out in the other answer, the eigenvalues of $M$ are $\lambda_k=1-2iv\sin\frac{2\pi k}{N}$, where $k=0,1,\ldots,\,N-1$. Therefore, when $N$ is even, $\lambda_0=\lambda_{N/2}=1$ is the only real eigenvalue of both $M$ and $M^{-1}$ and it has multiplicity $2$. However, by Perron-Frobenius theorem, if $M^{-1}$ is positive, it must possess a simple real eigenvalue (namely, the spectral radius $\rho(M^{-1})$).
