Alternative computation of eigenvalues of this tridiagonal matrix Consider the tridiagonal symmetric pd matrix 
$$ M=\begin{bmatrix} 2 & -1 &\dots \\ -1 & 2 &-1&\dots \\ \vdots & \ddots & \ddots & \ddots \\ 0 & \dots & -1 & 2 & -1 \\ 0  &\dots &\dots & -1 & 1\end{bmatrix}$$
The question is how to find eigenvalues and eigenvectors of $M$. I already know a way, which consists in short as introducing the sequence on the components $\phi_i$ of an eigenvector $\phi$ (the sequence is $-\phi_{k-1}+2\phi_k-\phi_{k+1}=\lambda \phi_k$). Then it is possible to find the general term and $\lambda$ using the polynomial associated with the sequence. After some manipulation, the final result is: 
$$\lambda_k=2\Big(1-\cos\Big(\dfrac{(2k-1)\pi}{2n+1}\Big)\Big)$$
and the eigenvectors are given by
$$\phi_k^{(i)}=\sin\Big(\dfrac{i(2k-1)\pi}{2n+1}\Big)$$
where $i$ is the component index and $k$ the index of the eigenvector.
I was just wondering if someone knew another way of finding the eigendecomposition.
 A: Partial answer: We can find the eigenvalues directly from the characteristic polynomial.
For simplicity, let $M_n$ be the $n$-dimensional version of $M$ and let $D_n(z)$ denote the characteristic polynomial $\det{(M_n - z I)}$ of $M_n$. If we for convenience define $D_0 = 1$ and $D_1 = 1 - z$, it shouldn't be too difficult to convince ourselves that the sequence of characteristic polynomials satisfies the recurrence
\begin{align}
D_0 &= 1 \\
D_1 &= 1-z \\
D_{n+2} &= (2-z)D_{n+1} - D_n, \quad n \geq 2.
\end{align}
The solution to the recurrence takes the form
$$
D_n(z) = c_+ r_+^n + c_- r_-^n,
$$
where $r_{\pm} = [(2-z) \pm \sqrt{(z-2)^2 - 4}]/2$ are the roots of the polynomial $x^2 - (2-z)x + 1$. Using the prescribed values of $D_0$ and $D_1$ we soon find that
$$
c_+ = \frac{1 - z - r_-}{r_+ - r_-}, \\
c_- = \frac{1 - z - r_+}{r_+ - r_-},
$$
such that
$$
D_n(z) = \frac{1 - z - r_-}{r_+ - r_-} r_+^n - \frac{1 - z - r_+}{r_+ - r_-} r_-^n.
$$
Now make the substitution $z = 2(1- \cos{(\phi)})$ (Note: we should really first check that the eigenvalues are real and confined to $0 < \lambda < 4$ to do this, but that shouldn't be too hard). After a little trigonometry this comes out as
$$
D_n(\phi) = 2 \frac{\sin{(\phi/2)}}{\sin{(\phi)}} \cos{\left( \left(n + \frac{1}{2} \right)\phi \right)}.
$$
Solving $D_n(\phi) = 0$ gives the desired
$$
\phi = \frac{(2k-1)\pi}{2n+1}, \quad k = 1, \dots, n,
$$
or
$$
\lambda_k = z = 2 \left( 1 - \cos{\left( \frac{(2k-1)\pi}{2n+1} \right)} \right).
$$
From here, I don't think there is a simpler way of finding the eigenvectors than the one you used, except that now we know $\lambda$ in advance.
A: Probably, an alternative way is to consider this matrix as an approximation up to a factor of the 1d laplacian with dirichlet boundary condition for x = 0 and neumann for x = 1. Then the eigenvector is a discrete version of a continuous eigenfunction which can be found by solving the corresponding differential eigenproblem
