Conditions for symmetric, Toeplitz $\mathbf{M}$ with nonnegative elements to have inverse with nonnegative elements Problem
Suppose we have symmetric, Toeplitz matrix $\mathbf{M}$ such that
$$
\mathbf{M} = \begin{bmatrix}
m_0 & m_1 & m_2 & m_3 & \cdots  &m_{n-1} \\
m_1 & m_0 & m_1 & m_2 & \cdots & m_{n-2} \\
m_2 & m_1 & m_0 & m_1 & \cdots & m_{n-3} \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
m_{n-1} & m_{n-2} & m_{n-3} & m_{n-4} & \cdots & m_0 \\
\end{bmatrix}
$$
where $m_0,m_1,\cdots,m_{n-1}$ all nonnegative, i.e. $m_i \ge 0, \forall i=0,1,\cdots,n-1$  .

What are the conditions for $\mathbf{M}^{-1}$ to have nonnegative elements?

The above problem assumes $\mathbf{M}$ is invertible (related).

Some notes

*

*A very simple example may be an identity matrix $I$, with $I^{-1} = I$ has all nonnegative elements.


*A diagonal dominance $m_0 > \sum_{j\neq i} m_{ij}, \forall i$ may guarantee the existence of inverse, but not what I want. For instance, $\tilde{\mathbf{M}} := \begin{bmatrix} 1 & 1/2 \\ 1/2 & 1 \end{bmatrix}$  has inverse $\tilde{\mathbf{M}}^{-1} = \begin{bmatrix} 4/3 & -2/3 \\ -2/3 & 4/3 \end{bmatrix}$.

Try
A paper gives an explicit form of the inverse of $\mathbf{M}$, but it does not help for finding an explicit representation.
Although it seems evident that if $m_0 > 0$ the diagonal of $\mathbf{M}^{-1}$ may be all positive, I think off-diagonals are not quite straight-forward. Any help will be appreciated.
 A: In general, if $A$ and $B$ are two nonnegative $n\times n$ matrices such that $AB=I$, one can prove by mathematical induction on $n$ that the sign patterns of $A$ and $B$ must be some permutation matrices (that are transposes of each other).
In the inductive step, since $1=\delta_{11}=\sum_ka_{ik}b_{ki}$, both $a_{1k}$ and $b_{k1}$ must be positive for some $k$. By permuting the columns of $A$ and the rows of $B$ if necessary, we may assume that $a_{11}$ and $b_{11}$ are positive. As
$$
\begin{cases}
0=\delta_{1j}=\sum_k a_{1k}b_{kj}\ge a_{11}b_{1j}\ge0,\\
0=\delta_{i1}=\sum_k a_{ik}b_{k1}\ge a_{i1}b_{11}\ge0,
\end{cases}\tag{1}
$$
we see that $a_{i1}=b_{1j}=0$ for all $i,j\ne1$. However, as $BA$ is also equal to $I$, if we interchange the roles of $A$ and $B$ in $(1)$, we also have $b_{i1}=a_{1j}=0$ for all $i,j\ne1$. Therefore $A=a_{11}\oplus X$ and $B=b_{11}\oplus Y$ for some $X$ and $Y$ with $XY=I$. By induction assumption, the sign patterns of $X$ and $Y$ are permutation matrices. Hence the sign patterns of $A$ and $B$ are permutation matrices too.
So, in your case, the sign pattern of $M$ must be a permutation matrix. As $M$ itself is a symmetric Toeplitz matrix, it must be in the form of $mI_n$ when $n$ is odd. When $n$ is even, $M$ must take the form of either $mI_n$ or $\pmatrix{0&mI_{n/2}\\ mI_{n/2}&0}$.
