# Jacobi Matrix

1. Jul 26, 2009

### dracolnyte

1. The problem statement, all variables and given/known data
An n x n array Hn = (hij) is said to be a jacobi matrix if hij = 0 whenever |i - j| >= 2. Suppose Hn also has the property that for each index i, hii = a, hi, i+1 = b and hi,i-1 = c. For instance, H4 =

a b 0 0
c a b 0
0 c a b
0 0 c a

(i) Show that det Hn = a (det Hn-1) - bc (det Hn-2) for n = 3,4,....

(ii) Find det H6.

3. The attempt at a solution
I was thinking of converting this into an upper triangle and use that theorem to solve by finding the product of the main diagonals. but I don't know how to prove for general n x n case and I don't know what it means by det Hn-1 and Hn-2

2. Jul 26, 2009

### HallsofIvy

Staff Emeritus
Looks like a straightforward calculation. Hn-1 and Hn-2 are, of course, the Jacobi matrices of order n-1 and n-2.

Expand Hn on the first row. Then Hn= a(cofactora)- b(cofactorb). It should be easy to see that "cofactora" is just Hn-1, so that part is aHn-1 while "cofactorb" has first column consisting of "c 0 0 ...". Expand that on the first column and you have "bcHn-2".

For example, with n= 4 you have
$$H_3= \left|\begin{array}{cccc}a & b & 0 & 0 \\ c & a & b& 0 \\ 0 & c & a & b \\0 & 0 & c & a\end{array}\right|= a\left|\begin{array}{ccc}a & b & 0 \\ c & a & b \\ 0 & c & b\end{array}\right|- b\left|\begin{array}{ccc}c & b & 0 \\ 0 & a & b \\ 0 & c & a\end{array}\right|$$
$$= a\left|\begin{array}{ccc}a & b & 0 \\ c & a & b \\ 0 & c & b\end{array}\right|- bc\left|\begin{array}{cc}a & b \\ c & a \end{array}\right|= aH_2- bcH_1$$