Question

Show that the \(n\)-rowed determinant $$ \left|\begin{array}{ccccccc} \cos \theta & 1 & 0 & 0 & & & 0 \\ 1 & 2 \cos \theta & 1 & 0 & & \vdots & 0 \\ 0 & 1 & 2 \cos \theta & 1 & & & 0 \\ 0 & 0 & 1 & 2 \cos \theta & & & 0 \\ & & \cdots & & & & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & 2 \cos \theta \end{array}\right|=\cos n \theta . $$

Short answer

The determinant of the n-rowed matrix is \( \text{cos}(n\theta) \).

Step-by-step solution

Define the determinant

Let the determinant be denoted by \(D_n\), where \(D_n\) is the determinant of the given \(n\)-rowed matrix.

Check the base cases

Calculate \(D_1\) and \(D_2\) as the base cases to confirm the pattern. For \(D_1\):\[ D_1 = \begin{vmatrix} \text{cos}\theta \end{vmatrix} = \text{cos} \theta \] This matches with \( \text{cos}(1\theta) \). For \(D_2\):\[ D_2 = \begin{vmatrix} \text{cos}\theta & 1 \ 1 & 2 \text{cos}\theta \end{vmatrix} \ D_2 = \text{cos}\theta(2\text{cos}\theta) - 1(1) \ D_2 = 2\text{cos}^2\theta - 1 \ D_2 = \text{cos}(2 \theta) \] Both base cases confirm the desired result.

Develop the recurrence relation

Assume \(D_n\) is true and express \(D_n\) in terms of \(D_{n-1}\) and \(D_{n-2}\):\[ D_n = \text{cos}\theta (\text{det of } (n-1)\text{-row matrix}) - (\text{det of } (n-2)\text{-row matrix}) \]By minor expansion on the first row:\[ D_n = \text{cos}\theta D_{n-1} - D_{n-2} \]

Use trigonometric identities

Recall the trigonometric identity for cosine of sum:\[ \text{cos}(n\theta) = 2 \text{cos}(\theta) \text{cos}((n-1)\theta) - \text{cos}((n-2)\theta) \]By induction, assume \(D_k = \text{cos}(k\theta)\) for all \(k < n\). Then using the recurrence relation:\[ D_n = 2 \text{cos} \theta \text{cos} ((n-1)\theta) - \text{cos}((n-2)\theta) \]This matches the trigonometric identity.

Conclude by induction

By mathematical induction, the pattern holds for all \(n\). Therefore, the n-rowed determinant equals \( \text{cos}(n\theta) \).

Key concepts

Determinant Calculation

In linear algebra, a determinant is a scalar value that can be computed from the elements of a square matrix. The determinant provides important properties of the matrix, especially regarding whether a matrix is invertible. For an \(n\) x \(n\) matrix, the determinant can be calculated using various methods, including cofactor expansion. Each element of the matrix contributes to the overall value of the determinant, influenced by their position and the surrounding elements.
In the given problem, our goal is to show that the determinant of the special tridiagonal matrix equals \(\text{cos}(n \theta) \). We start by breaking down the matrix into smaller matrices for easier calculation then progressively use those simplified forms to prove the general formula.

Recurrence Relations

Recurrence relations are equations that define sequences of values using recurrence. These sequences often rely on previous terms to establish the subsequent terms.
In this exercise, the key recurrence relation helps to connect the determinants of matrices of different sizes. Specifically, we formulate that \(D_n \) depends on \((n-1)\)-rowed and \((n-2)\)-rowed matrices:
\[ D_n = \text{cos} \theta D_{n-1} - D_{n-2} \]
This relationship simplifies the recursive calculation of the determinant, ensuring we can build each subsequent term upon the previous ones, ultimately leading to the desired result.

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish the truth of an infinite sequence of statements. We leveraged induction in this exercise to prove the determinant formula for matrices of any size.
The process includes:

• Base Case: Verify the formula holds for the smallest values of \(n\). We computed \(D_1\) and \(D_2\), which correctly matched the given formula \(\text{cos}(n \theta)\).
• Inductive Step: Assume \(D_k = \text{cos}(k \theta)\) holds for all values less than some integer \(n\), then show that \(D_n\) also holds. By using the recurrence relation and trigonometric identities, we demonstrated that the induction hypothesis propels the pattern for all integers \(n\).

This method provides a rigorous proof that the determinant indeed follows the pattern predicted.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. In this exercise, the critical trigonometric identity used is:
\[ \text{cos}(n \theta) = 2 \text{cos}(\theta) \text{cos}((n-1) \theta) - \text{cos}((n-2) \theta) \]
This identity, known as the cosine multiple-angle formula, relates the cosine of multiple angles to the cosines of smaller angles. We applied this identity to connect our recurrence relation \(D_n = \text{cos} \theta D_{n-1} - D_{n-2} \) throughout the induction proof.
By tying the recurrence relation with the trigonometric identity, we highlighted the equivalence of the determinant results to the trigonometric expression, validating that \(D_n = \text{cos}(n \theta)\).