Derivation of the Determinant of a Tridiagonal Matrix
Formula
$$ X_{n} := \begin{bmatrix} x & 1 & 0 & \cdots & 0 & 0 \\ 1 & x & 1 & \cdots & 0 & 0 \\ 0 & 1 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x & 1 \\ 0 & 0 & 0 & \cdots & 1 & x \end{bmatrix} $$ The determinant of the tridiagonal matrix $X_{n}$ defined as above is given by $$ \det X_{n} = \left| X_{n} \right| = U_{n} \left( {{x} \over {2}} \right) $$
$U_{n}$ denotes the Chebyshev polynomial of the second kind of degree $n$.
Proof
Of course, $X_{n}$ is not a general tridiagonal matrix but a special form even among tridiagonal Toeplitz matrices. In particular, it is a form usefully employed in the numerical solution of partial differential equations, and one is usually most interested in its eigenvalues.
Part 1. $| X_{n+1} | = x | X_{n} | - | X_{n-1} |$
Laplace expansion: with respect to a chosen row $i$, $$ \det A = \sum_{i=1}^{n} a_{ij} C_{ij} $$
Let us define the $n \times n$ matrix $Y_{n} : = \begin{bmatrix} 1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & x & 1 & \cdots & 0 & 0 \\ 0 & 1 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x & 1 \\ 0 & 0 & 0 & \cdots & 1 & x \end{bmatrix}$. The first column of $Y_{n}$ is $(1, 0 , \cdots , 0)$, which becomes the zero vector once the first component is excluded, so the cofactors involving the first column are $0$. Therefore,
$$ | Y_{n} | = 1 \cdot | X_{n-1} | - 1 \cdot 0 + 0 \cdot 0 - \cdots \pm 0 \cdot 0 = | X_{n-1} | $$
Similarly, by the Laplace expansion,
$$ | X_{n+1} | = x \cdot | X_{n} | - 1 \cdot | Y_{n} | + 0 \cdot 0 - \cdots \pm 0 \cdots 0 = x | X_{n} | - | X_{n-1} | $$
Part 2. $\displaystyle | x_{n}| = U_{n} \left( {{x} \over {2}} \right)$
Define the proposition $\displaystyle P(n) : | x_{n}| = U_{n} \left( {{x} \over {2}} \right)$ and use mathematical induction.
When $n=1$, $$ | X_{1} | = x \\ U_{1} \left( {{x} \over {2}} \right) = 2 {{x} \over {2}} = x $$
When $n=2$, $$ | X_{2} | = x^2 - 1 \\ U_{2} \left( {{x} \over {2}} \right) = 4 \left( {{x} \over {2}} \right)^2 - 1 = x^2 - 1 $$
Assume that the proposition $P(n)$ holds for $n=k-1$ and $n=k$.
Recurrence relation of the Chebyshev polynomial of the second kind:
$$ U_{n+1} (x) = 2x U_{n} (x) - U_{n-1} (x) $$
From Part 1,
$$ | X_{k+1} | = x | X_{k} | - | X_{k-1} | $$
and
$$ U_{k+1} \left( {{x} \over {2}} \right) = x U_{k} \left( {{x} \over {2}} \right) - U_{k-1} \left( {{x} \over {2}} \right) $$
Therefore,
$$ | X_{k+1} | - U_{k+1} \left( {{x} \over {2}} \right) = x \left[ | X_{k} | - U_{k} \left( {{x} \over {2}} \right) \right] - \left[ | X_{k-1} | - U_{k-1} \left( {{x} \over {2}} \right) \right] $$
By the assumption, $\displaystyle | X_{k} | = U_{k} \left( {{x} \over {2}} \right)$ and $\displaystyle | X_{k-1} | = U_{k-1} \left( {{x} \over {2}} \right)$, so
$$ | X_{k+1} | - U_{k+1} \left( {{x} \over {2}} \right) = 0 $$
By mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$.
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