Determinant of a tridiagonal Toeplitz matrix

Question

Let $A_n$ denote an $n \times n$ tridiagonal matrix. $$A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2$$

Set $D_n = \det(A_n)$. Prove that $D_n = 2D_{n-1} - D_{n-2}$ for $n \ge 4$.

Answer 1

Expand with respect to the first column: you get $$ D_n =2 \times \begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)} - (-1) \times \begin{vmatrix}-1 & 0 & & & 0 \\ -1 & 2 & -1 & & 0 \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)} $$

Now expand according to the first line the second determinant: you get $$ =2 \times \begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)} - (-1)^2\times \begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-2)} $$ Conclusion: $$ D_n = 2D_{n-1} - D_{n-2} $$

Answer 2

Here is a way to compute the determinant without using induction. Multiply the matrix $A_n$ to the left by the $n\times n$ upper triangular matrix $U_n$ with all entries on and above the diagonal equal to$~1$: $$ \begin{pmatrix}1&1&1&\ldots&1\\ 0&1&1&\ddots&\vdots\\0&0&1&\ddots&1\\ \vdots & \ddots & \ddots & \ddots & 1 \\0&0& \ldots &0 & 1 \end{pmatrix} \begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} = \begin{pmatrix}1&0&0&\ldots&0&1\\ -1&1&0&\ddots&0&1\\0&-1&1&\ddots&0&1\\ \vdots & \ddots & \ddots & \ddots & 0&1 \\ \vdots & \ddots & \ddots & -1 & 1&1 \\0&0& \ldots &0&-1 & 2 \end{pmatrix}. $$ Since $\det(U_n)=1$, we shall get $\det(A_n)$ as the determinant of our resulting matrix$~R$.

Now recognise that $R=I-C_P$ where $C_P$ is the companion matrix of the polynomial $P=X^n+X^{n-1}+\cdots+X^2+X^1+X^0$. Since it is well known that $P$ is the characteristic polynomial $\det(IX-C_p)$ of $C_P$ we get $$ \det(A_n)=\det(R)=\det(I-C_P)=\det(IX-C_P)[X:=1]=P[X:=1]=n+1. $$