Suppose I have a $n\times n$ symmetric tridiagonal matrix: $$ A = \left(\begin{array}{ccccc} 1 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & & \cdots \\ 0 & -1 & 2 & \cdots & 0 \\ \cdots & & \cdots & 2 & -1 \\ 0 & \ldots & 0 & -1 & 1 \end{array}\right) $$ I know that the smallest eigenvalue of the matrix $A$ is 0. I want to know if there exists a constant $c$ such that the smallest positive (i.e., the second smallest) eigenvalue of the matrix $A$ is larger than $c/n$.
By now, I prove that the smallest positive eigenvalue of $A$ is larger than $1/n(n+1)$. Let $$ B = \left(\begin{array}{ccccc} 1 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & & \cdots \\ 0 & -1 & 2 & \cdots & 0 \\ \cdots & & \cdots & 2 & -1 \\ 0 & \ldots & 0 & -1 & 2 \end{array}\right)\in \mathbb{R}^{n-1 \times n-1} $$ be the $n$-th principal submatrix of $A$, then we know that the smallest positive eigenvalue of $A$ is larger than the smallest eigenvalue of $B$, by the seperation theorem.
According to DOI 10.1515/math-2016-0091, the smallest eigenvalue of $B$ is larger than $$ \|B^{-1}\|_{\infty}=1/n(n+1). $$
The answer is no. In fact, the smallest non-zero eigenvalue is bounded above by $c/n^2$ for a suitable $c>0$. With your lower bound, we can deduce that this eigenvalue is asymptotically $\Theta(1/n^2)$.
Note that $-A$ is the matrix associated with the Discretized second derivative, taking Neumann boundary conditions and a step-size of $h=1$. Thus, the eigenvalues of $A$ are given by $$ \lambda_j = 4 \sin^2 \left(\frac{\pi j}{2n} \right), \quad j = 0,1,\dots,n-1. $$ Conveniently, we have $\lambda_0 \leq \lambda_1 \leq \cdots \leq \lambda_{n-1}$. So, the smallest non-zero value among these eigenvalues is given by $\lambda_1 = 4\sin^2(\pi/(2n)) \leq 4\left(\frac{\pi}{2n}\right)^2 = c \cdot \frac 1{n^2}$, where $c = \pi^2$.