Tri-diagonal Matrix Determinant Derivation
Formulas
For the tridiagonal matrix $X_{n} := \begin{bmatrix} x & 1 & 0 & \cdots & 0 & 0 \\ 1 & x & 1 & \cdots & 0 & 0 \\ 0 & 1 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x & 1 \\ 0 & 0 & 0 & \cdots & 1 & x \end{bmatrix}$, $$ | x_{n}| = U_{n} \left( {{x} \over {2}} \right) $$
$U_{n}$ denotes the second-kind Chebyshev polynomials of degree $n$.
Of course, $X_{n}$ is not just any tridiagonal matrix but a special form of tridiagonal Toeplitz matrices. Among them, it is particularly used in the numerical solutions of partial differential equations, with usual interest in its eigenvalues.
Proof
**Part 1. $| X_{n+1} | = x | X_{n} | - | X_{n-1} |$
Laplace expansion: For the selected row $i$, $$ \det A = \sum_{i=1}^{n} a_{ij} C_{ij} $$
Let’s define matrix $n \times n$ for $Y_{n} : = \begin{bmatrix} 1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & x & 1 & \cdots & 0 & 0 \\ 0 & 1 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x & 1 \\ 0 & 0 & 0 & \cdots & 1 & x \end{bmatrix}$. The first column of $Y_{n}$, being $(1, 0 , \cdots , 0)$, becomes a zero vector except for its first component, which makes the cofactor including the first column $0$. Hence,
$$ | Y_{n} | = 1 \cdot | X_{n-1} | - 1 \cdot 0 + 0 \cdot 0 - \cdots \pm 0 \cdot 0 = | X_{n-1} | $$
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 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 $n=2$, $$ | X_{2} | = x^2 - 1 \\ U_{2} \left( {{x} \over {2}} \right) = 4 \left( {{x} \over {2}} \right)^2 - 1 = x^2 - 1 $$
Assuming proposition $P(n)$ holds for $n=k-1$ and $n=k$,
Recurrence formula of the second-kind Chebyshev polynomials:
$$ U_{n+1} (x) = 2x U_{n} (x) - U_{n-1} (x) $$
From Part 1,
$$ | X_{k+1} | = x | X_{k} | - | X_{k-1} | $$
and,
$$ U_{k+1} \left( {{x} \over {2}} \right) = x U_{k} \left( {{x} \over {2}} \right) - U_{k-1} \left( {{x} \over {2}} \right) $$
Thus,
$$ | X_{k+1} | - U_{k+1} \left( {{x} \over {2}} \right) = x \left[ | X_{k} | - U_{k} \left( {{x} \over {2}} \right) \right] - \left[ | X_{k-1} | - U_{k-1} \left( {{x} \over {2}} \right) \right] $$
Given that $\displaystyle | X_{k} | = U_{k} \left( {{x} \over {2}} \right)$ and $\displaystyle | X_{k-1} | = U_{k-1} \left( {{x} \over {2}} \right)$,
$$ | X_{k+1} | - U_{k+1} \left( {{x} \over {2}} \right) = 0 $$
By mathematical induction, proposition $P(n)$ is true for $n \in \mathbb{N}$.
■