# Determinant of a tridiagonal Toeplitz matrix

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 $$n \ge 4$$.

• What have you tried so far? If you don't know where to begin, perhaps you should make use of the variable "n." Oct 28, 2014 at 22:39
• Feb 15, 2015 at 9:14
• For a more general version, see this answer. Feb 15, 2015 at 10:43
• I've "unclosed" this question as duplicate of math.stackexchange.com/q/266998/18880, because none of the answers to that question make it even remotely obvious that for this specific case the determinant is $n+1$. (I know this question does not explicitly ask to compute it, but another question marked as duplicate of this one does.) Much as I like closing for abstract duplicates, I think this question is better not marked as duplicate of a considerably harder question. Feb 15, 2015 at 11:30

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: $$D_n = 2D_{n-1} - D_{n-2}$$
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 $$\det(A_n)=\det(R)=\det(I-C_P)=\det(IX-C_P)[X:=1]=P[X:=1]=n+1.$$