Symmetric Tridiagonal Matrix has distinct eigenvalues. Show that the rank of $ n\times n$ symmetric tridiagonal matrix is at least $n-1$, and prove that it has $n$ distinct eigenvalues.

This is for tridiagonal matrices with nonzero off-diagonal elements.
Let $\lambda$ be an eigenvalue of $A\in\mathbb{R}^{n\times n}$ (which is symmetric tridiagonal with nonzero elements $a_{2,1},a_{3,2},\ldots,a_{n,n-1}$ on the subdiagonal). The submatrix constructed by deleting the first row and the last column of $A-\lambda I$ is nonsingular (since it is upper triangular and has nonzero elements on the diagonal) and hence the dimension of the nullspace of $A-\lambda I$ is 1 (because its rank cannot be smaller than $n-1$ and the nullspace must be nontrivial since $\lambda$ is an eigenvalue). It follows then that the geometric multiplicity is 1 and hence the algebraic multiplicity of $\lambda$ is 1 as well. This holds for any eigenvalue of $A$ and hence they are distinct.
The fact that $\mathrm{rank}(A)\geq n-1$ is just a simple consequence of that ($0$ has also multiplicity 1 if $A$ is singular).
