To simplify things a little, I describe conditions under which the smallest eigenvalue is strictly positive. These can be adjusted to get equality to zero.
Let $\alpha := a_j=a_k$ be the smallest elements on the diagonal. It can be shown that if \begin{equation*} \frac{2}{n(n-1)}\alpha^2 - \frac{1}{4}\sum_{i < j} (a_i-a_j)^2 < n-1, \end{equation*} then the tridiagonal matrix is positive definite.
If you want necessaryNecessary and sufficient conditions, then things for positive definiteness of the tridiagonal matrix in question are slightly more involveddescribed below.
Definition (Chain Sequence). A sequence $\lbrace x_k \rbrace_{k > 0}$ is a chain sequence if there exists another sequence $\lbrace y_k \rbrace_{k\ge 0}$ such that \begin{equation*} x_k = y_k(1-y_{k-1}), \end{equation*} where $y_0 \in [0,1)$ and $y_k \in (0,1)$ for $k > 0$.
By the Wall-Wetzel Theorem, your tridiagonal matrix is positive definite if and only if
\begin{equation*} \left\lbrace \frac{1}{a_ka_{k+1}} \right\rbrace_{k=1}^{n-1} \end{equation*}
is a chain sequence.
Example. In particular, if the entries of the matrix satisfy,
\begin{equation*} 0 < \frac{1}{a_ka_{k+1}} < \frac{1}{4\cos^2\left(\frac{\pi}{n+1}\right)},\quad k=1,\ldots,n-1, \end{equation*} then it is positive definite.
For additional information and details about this material, please see:
- M. Andelic, and C. M. Da Fonesca. Sufficient conditions for positive definiteness of tridiagonal matrices revisited. (2010).