5
$\begingroup$

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 $n \ge 4$.

$\endgroup$
4
  • 1
    $\begingroup$ What have you tried so far? If you don't know where to begin, perhaps you should make use of the variable "n." $\endgroup$
    – daOnlyBG
    Oct 28, 2014 at 22:39
  • $\begingroup$ See also: math.stackexchange.com/questions/1148581/… $\endgroup$ Feb 15, 2015 at 9:14
  • $\begingroup$ For a more general version, see this answer. $\endgroup$ Feb 15, 2015 at 10:43
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 15, 2015 at 11:30

2 Answers 2

6
$\begingroup$

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: $$ D_n = 2D_{n-1} - D_{n-2} $$

$\endgroup$
6
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .