Here is a Toeplitz matrix of the form:
\begin{pmatrix} a_{0} & a_1 & a_2 &\cdots&a_n\\ a_{-1} & a_0 & a_1 &\cdots&a_{n-1}&\\ a_{-2} & a_{-1} & a_0&\cdots&a_{n-2}\\ \vdots & \vdots & \vdots &\ddots&\vdots\\ a_{-n} & a_{-n+1} & a_{-n+2}&\cdots&a_0 \end{pmatrix}
We assume that the matrix $A \geq 0$, which means that $(AX, X) \geq 0$ $\forall X \in \mathbb{R}^{n+1}$. We also assume that the determinant of this matrix is $0$.
I know that the $\det = 0$ constraint the values of the elements down the diagonal like in a $2 \times 2$ matrix, the value of the element down the main diagonal should be the product of the element on the main diagonal divided by the other element, i.e $a_{2,1} = (a_{1,1}\times a_{2,2}) \div a_{1,2} $
Since $A$ is not invertible ($\det A=0$) then there must be a vector $x$ such that $x^TA=0$ (this follows since $A^T$ is also singular). Expanding out $$0=x^TA=\sum_{i=1}^nx_i\mathbf{a}_i$$ where $\mathbf{a}_i$ is the $i$th row of $A$; so if $x_1\ne0$ then $\mathbf{a}_1$ can be written as a linear combination of the rest of the rows.
However this is not always possible. For example, $\begin{pmatrix}4&8&5\\2&4&8\\1&2&4\end{pmatrix}$ is of the given form and singular, yet the first row cannot be written as a linear combination of the other rows.