Eigenvalues of a symmetric Jacobi matrix

Let $A \in M_n(\mathbb{R})$ be the symmetric Jacobi matrix defined by $A=\begin{pmatrix} 2/h & -1/h & & \huge0 \\ -1/h & \ddots & \ddots \\ & \ddots & \ddots & -1/h\\ \huge0 & &-1/h & 2/h \end{pmatrix}$.

I would like to compute the eigenvalues (and the eigenvectors) of this matrix, for a fixed $n>0$. So far, I've tried two different ways without success, namely I've shown the classical recurrence relation of such a determinant and then 1) tried to solve the recurrence relation of the characteristic polynomial, 2) tried to find a pattern of the eigenvalues for small $n$'s.

In particular, if my computations are correct, the first few eigenvalues are:

1. $\frac{2}{h} \qquad$ 2. $\frac{1}{h},\frac{3}{h} \qquad $ 3. $ \frac{2}{h}, \frac{2 \pm \sqrt{2}}{h} \qquad$ 4. $\frac{3 \pm \sqrt{5}}{2h}, \frac{5 \pm \sqrt{5}}{2h} \qquad$ 5. $\frac{1}{h},\frac{2}{h},\frac{3}{h},\frac{2 \pm \sqrt{3}}{h}$.

Can someone help me, please?

A simplification is brought by considering $hA$ which is a Toeplitz triadiagonal matrix with eigenvalues given by formulas :

$$\lambda_k=a+2 \sqrt{bc} \cos \frac{k \pi}{n+1}\tag{1}$$

with $a=2$ on the main diagonal and $b = c = -1$ on the two other sub-diagonals, giving:

$$\lambda_k=2+2 \cos \frac{k \pi}{n+1}\tag{1}$$

as given here.

These are the eigenvalues of $hA$.

Therefore the eigenvalues of $A$ are obtained by dividing all of them by $h$ giving formula :

$$\lambda'_k=\frac{2}{h}(1+\cos \frac{k \pi}{n+1})\tag{2}$$

which are indeed all positive, showing that matrix $A$ is positive definite.

You can check that the eigenvalues you have found correspond to formula (1).