Uniqueness of solution for a tridiagonal system I have a claim I’ve been conjecturing. Not sure if it’s true or not. Context: I’m doing some calculations with finite difference schemes.
Say I have the following real $n \times n$ tridiagonal matrix $A$:
$$
\begin{bmatrix}
\;\;\;2 & -1 & & &\\
-1 & \;\;\;2 & -1 & &\\
 & \ddots & \ddots & \ddots\\
 & & -1 & \;\;\;2 & -1 \\
 & & & -1 & \;\;\;2 
\end{bmatrix}
$$
Does the following system have a unique solution?
$A\mathbf{U}=\mathbf{F}$ where $\mathbf{U} = \begin{bmatrix}
U_{1}\\ 
U_{2}\\ 
\vdots\\ 
\\
U_{n} 
\end{bmatrix}$ and $\mathbf{F}$ is just a real vector of dimension $n$.
Observations:
Other than the fact it’s tridiagonal, I noticed that it is diagonally dominant. I also think I can compute an $LU$ factorization, but I’m not sure how that would help. Any directions?
 A: Rodrigo is perfectly right, but we may prove that $A_n$ (the $n\times n$ matrix with such a structure) is invertible by simply computing its determinant through a recursive approach. An expansion along the last row gives:
$$\det A_n = 2\cdot\det A_{n-1}-\det A_{n-2} \tag{1}$$
hence:
$$ \det A_n = Cn +D \tag{2}$$
and since $\det A_1=2$ and $\det A_2=3$, $\color{red}{\det A_n = n+1\neq 0}$ and $A_n$ is invertible.
$(2)$ follows from the fact that the characteristic polynomial of the sequence $\{\det A_n\}_{n\geq 1}$ is $p(x)=(x-1)^2$ by $(1)$.
A: $\mathrm A$ is not merely tridiagonal, it is also Toeplitz. Hence, the $n$ real eigenvalues of $\mathrm A$ are given by [0]
$$\lambda_k (\mathrm A) = 2 + 2 \cos \left(\frac{k \pi}{n+1}\right)$$
for $k \in \{1,2,\dots,n\}$. Thus,
$$0 < 2 - 2 \cos \left(\frac{\pi}{n+1}\right) \leq \lambda_k (\mathrm A) \leq 2 + 2 \cos \left(\frac{\pi}{n+1}\right) < 4$$
and we conclude that $\mathrm A$ is invertible.

[0] Silvia Noschese, Lionello Pasquini, and Lothar Reichel, Tridiagonal Toeplitz Matrices: Properties and Novel Applications, 2006.
A: Note that
$$\mathrm x^T \mathrm A \mathrm x = x_1^2 + \underbrace{\left(\sum_{i=1}^{n-1} (x_{i+1} - x_i)^2\right)}_{=: f(x)} + x_n^2$$
where $f$ is positive semidefinite and vanishes on $\{\gamma 1_n \mid \gamma \in \mathbb R\}$. Fortunately, $x_1^2 + x_n^2$, which is also positive semidefinite, does not vanish on $\{\gamma 1_n \mid \gamma \in \mathbb R\}$, which allows us to conclude that $\mathrm A$ is positive definite and, thus, invertible.
