Inverse of nonnegative Toeplitz matrice Consider a right-hand circulant matrice of size $n$ (called also Toeplitz matrice) \begin{equation} T= \left( \begin{array}{ccccc} a_1 & a_2 & a_3 & \ldots & a_n \\\ a_n & a_1 & a_2 & \ldots & a_{n-1}\\\ \vdots\\\ a_2 & a_3 & a_4 & \ldots & a_1 \end{array} \right) \end{equation} $\\{a_i\\}$ are nonnegative and not all equal. Is $T$ alwayls invertible? I proved that: $\bullet$ for n=2 the $Det(T) = (a_1+a_2)(a_1-a_2) \neq 0$ $ $\bullet$ for n=3 the $Det(T) = \frac{1}{2}(a_1+a_2+a_3)\left( (a_1-a_2)^2 + (a_1-a_3)^2(a_2-a_3)^2 \right) \neq 0$ What about $n>3$?

a long hint: does writing $T$ as a linear combination of the powers of $P $ where $p_{i i-1} = 1$ and zero every where else. $P$ has eigenvalues the $n$th roots of unity. so $T$ thaw $a_1+a_2\omega+a_3\omega^2+\cdots$ and the determinant of $T$ is the product of $(a_1+a_2\omega+a_3\omega^2+\cdots)(a_1+a_2\omega+a_3\omega^2+\cdots))\cdots$

showing that the determinant is nonzero reduces to showing that all the factors are nonzero. that is if $\omega$ in an $n$th root of unity and $a_1, a_2, \cdots$ are non negative and not all equal, then $$(a_1+a_2\omega+a_3\omega^2+\cdots )\
eq 0$$

it is more likely to be true if $n$ is a prime number. i am not sure if it is true for nonprime numbers. e.g., take $n = 4, \omega = i, a_1 = a2 = -a_3 = -a_4 = 1 $