# Proving that a symmetric tridiagonal matrix is positive definite

How to prove that the square and symmetric matrix below is positive definite?

$$A = \begin{bmatrix} 81 & -40 & \\ -40 & 101 & -40 & \\ 0 & -40 & 101 & -40 & \\ 0 & 0 & -40 & 101 & -40 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 101& -40\\ \dots & \dots & \dots & \dots & \dots & \dots & -40 & 20 \end{bmatrix}$$

I am self-studying linear algebra, and I wonder if I could prove this to be positive definite without using advanced theorems like the Gershgorin and Sylvester's criterion. What I know is, for a matrix to be positive definite, all eigenvalues must be positive or there is a vector $$x$$ such that $$x^T Ax > 0$$, where $$x^T$$ refers to the transposition of $$x$$. If I replace $$-40$$ by $$0$$, then $$A$$ is indeed positive definite since the diagonal entries are all positive. How could I continue this?

Note that $$100 = 80 + 20$$ and $$\sqrt{80} \cdot \sqrt{20} = 40$$. We have
\begin{aligned} x^{\mathsf{T}} A x &= 81 x_1^2 + 20 x_n^2 + 101 \sum_{1 < i < n} x_i^2 - 80 \sum x_i x_{i + 1} \\ &= \sum (\sqrt{80} x_i - \sqrt{20} x_{i + 1})^2 + \sum_{1 \le i < n} x_i^2 \\ &\ge 0 \text{.} \end{aligned}
And if the RHS $$= 0$$, then $$x_i = 0$$ for all $$i \in [1, n - 1]$$ and hence $$x_n$$, too.