For $n \geq 3$ find the eigenvalues (with their multiplicities) of the $n \times n$ matrix

$$\begin{bmatrix}
1 & 0 & 1 & 0 & 0 & 0 & \dots & \dots & 0 & 0\\ 
0 & 2 & 0 & 1 & 0 & 0 & \dots & \dots & 0 & 0\\ 
1 & 0 & 2 & 0 & 1 & 0 & \dots & \dots & 0 & 0\\ 
0 & 1 & 0 & 2 & 0 & 1 & \dots & \dots & 0 & 0\\ 
0 & 0 & 1 & 0 & 2 & 0 & \dots & \dots & 0 & 0\\ 
0 & 0 & 0 & 1 & 0 & 2 & \dots & \dots & 0 & 0\\ 
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots &  & \vdots & \vdots\\ 
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots &  & \ddots & \vdots & \vdots\\ 
0 & 0 & 0 & 0 & 0 & 0 & \dots & \dots & 2 & 0\\ 
0 & 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & 1
\end{bmatrix}$$
To find the eigenvalues (with their multiplicities) of the given \( n \times n \) matrix, we will follow these steps:

1. **Define the Matrix and Eigenvector:**
   Let the matrix \( A \) be:
   \[
   A = \begin{bmatrix}
   1 & 0 & 1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\
   0 & 2 & 0 & 1 & 0 & 0 & \dots & \dots & 0 & 0 \\
   1 & 0 & 2 & 0 & 1 & 0 & \dots & \dots & 0 & 0 \\
   0 & 1 & 0 & 2 & 0 & 1 & \dots & \dots & 0 & 0 \\
   0 & 0 & 1 & 0 & 2 & 0 & \dots & \dots & 0 & 0 \\
   0 & 0 & 0 & 1 & 0 & 2 & \dots & \dots & 0 & 0 \\
   \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots &  & \vdots & \vdots \\
   \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &  & \ddots & \vdots & \vdots \\
   0 & 0 & 0 & 0 & 0 & 0 & \dots & \dots & 2 & 0 \\
   0 & 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & 1
   \end{bmatrix}
   \]
   Let an eigenvector be \( \mathbf{x} = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^T \) with eigenvalue \( \lambda \).

2. **Set Up the Recurrence Relation:**
   For a generic \( k \), the recurrence relation is:
   \[
   x_{k-2} + 2x_k + x_{k+2} = \lambda x_k
   \]
   For \( k = 2 \), we have:
   \[
   2x_2 + x_4 = \lambda x_2 \quad \text{(set \( x_0 = 0 \) to fit the pattern)}
   \]
   For \( k = 1 \), we have:
   \[
   x_1 + x_3 = \lambda x_1 \quad \text{(set \( x_{-1} = -x_1 \) to fit the pattern)}
   \]
   Similarly, at the other end, set \( x_{n+1} = 0 \) and \( x_{n+2} = -x_n \).

3. **Extend to a Doubly Infinite Sequence:**
   Extend the sequence to hit zero and cross sign at \( 0 \) and \( n+1 \), so we should do so at every multiple of \( n+1 \).

4. **Solve the Recurrence Relation:**
   The recurrence relation \( x_{k-2} + 2x_k + x_{k+2} = \lambda x_k \) can be solved using characteristic equations. The characteristic equation is:
   \[
   \omega^{-2} + 2 + \omega^2 = (\omega + \omega^{-1})^2 = \lambda
   \]
   Since the original matrix is real symmetric, we are looking for real \( \lambda \). The solutions are linear combinations of \( \omega^k \) for the four solutions of the characteristic equation.

5. **Find the Eigenvalues:**
   Let \( \omega = e^{i\theta} \) with \( \theta = \frac{\pi j}{n+1} \cdot j \). The sequence \( x_k = \frac{1}{2i} (\omega^k + \omega^{-k}) = \sin(k\theta) \) is a solution to the recurrence with:
   \[
   \lambda = 4 \cos^2 \theta
   \]
   Similarly, \( \omega = -e^{-i\theta} \) leads to another solution \( x_k = (-1)^k \sin(k\theta) \) with the same \( \lambda \).

6. **Determine the Eigenvectors and Eigenvalues:**
   For \( j = 1, 2, \cdots, n \), \( x_k = \sin\left(k \cdot \frac{\pi j}{n+1}\right) \) is an eigenvector with eigenvalue:
   \[
   \lambda = 4 \cos^2 \frac{\pi j}{n+1}
   \]
   Going from \( j \) to \( n-j \) is the sign flip noted above, and this is a full set of \( n \) eigenvectors. Each eigenvalue has multiplicity \( 2 \), except for the zero eigenvalue with eigenvector \( \begin{bmatrix} 1 & 0 & -1 & 0 & -1 & \cdots \end{bmatrix} \) in the case that \( n+1 \) is even.

The final answer is \( \boxed{ \lambda = 4 \cos^2 \frac{\pi j}{n+1} } \) for \( j = 1, 2, \cdots, n \).