📂Matrix Algebra

Tri-diagonal Matrix Determinant Derivation

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 the following.
$$\det X_{n} = \left| X_{n} \right| = U_{n} \left( {{x} \over {2}} \right)$$

$U_{n}$ represents the Chebyshev polynomials of the second kind of degree $n$.

Proof

It should be noted that $X_{n}$ is not just any tridiagonal matrix but a specific form known as the tridiagonal Toeplitz matrix. This form is particularly useful in the numerical solution of partial differential equations, and the eigenvalues are often of primary interest.

Part 1. $| X_{n+1} | = x | X_{n} | - | X_{n-1} |$

Laplace expansion: With respect to the chosen $i$ row
$$\det A = \sum_{i=1}^{n} a_{ij} C_{ij}$$

Matrix $n \times n$ $Y_{n} : = \begin{bmatrix}
1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & x & 1 & \cdots & 0 & 0 \\ 0 & 1 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x & 1 \\ 0 & 0 & 0 & \cdots & 1 & x \end{bmatrix}$을 정의하자.$Y_{n}$의 첫번째 열은$(1, 0 , \cdots , 0)$으로써, 첫번째 성분을 제외하면 영벡터가 되므로 첫번째 열을 포함한 여인자는$0$ holds. Therefore,

$$| Y_{n} | = 1 \cdot | X_{n-1} | - 1 \cdot 0 + 0 \cdot 0 - \cdots \pm 0 \cdot 0 = | X_{n-1} |$$

Similarly, by Laplace expansion,

$$| X_{n+1} | = x \cdot | X_{n} | - 1 \cdot | Y_{n} | + 0 \cdot 0 - \cdots \pm 0 \cdots 0 = x | X_{n} | - | X_{n-1} |$$

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Part 2. $\displaystyle | x_{n}| = U_{n} \left( {{x} \over {2}} \right)$

Let’s define proposition $\displaystyle P(n) : | x_{n}| = U_{n} \left( {{x} \over {2}} \right)$ and apply mathematical induction.

For $n=1$,

$$| X_{1} | = x \\ U_{1} \left( {{x} \over {2}} \right) = 2 {{x} \over {2}} = x$$

For $n=2$,

$$| X_{2} | = x^2 - 1 \\ U_{2} \left( {{x} \over {2}} \right) = 4 \left( {{x} \over {2}} \right)^2 - 1 = x^2 - 1$$

Assume the proposition $P(n)$ holds for $n=k-1$ and $n=k$.

Recurrence relation of Chebyshev polynomials of the second kind:

$$U_{n+1} (x) = 2x U_{n} (x) - U_{n-1} (x)$$

From Part 1,

$$| X_{k+1} | = x | X_{k} | - | X_{k-1} |$$

implying,

$$U_{k+1} \left( {{x} \over {2}} \right) = x U_{k} \left( {{x} \over {2}} \right) - U_{k-1} \left( {{x} \over {2}} \right)$$

Hence,

$$| X_{k+1} | - U_{k+1} \left( {{x} \over {2}} \right) = x \left[ | X_{k} | - U_{k} \left( {{x} \over {2}} \right) \right] - \left[ | X_{k-1} | - U_{k-1} \left( {{x} \over {2}} \right) \right]$$

From the assumption, $\displaystyle | X_{k} | = U_{k} \left( {{x} \over {2}} \right)$ and $\displaystyle | X_{k-1} | = U_{k-1} \left( {{x} \over {2}} \right)$ imply

$$| X_{k+1} | - U_{k+1} \left( {{x} \over {2}} \right) = 0$$

By mathematical induction, $P(n)$ is true for $n \in \mathbb{N}$.

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