Solving Jacobi Matrices: Finding Det H6

[dracolnyte]

Homework Statement

An n x n array H_n = (h_ij) is said to be a jacobi matrix if h_ij = 0 whenever |i - j| >= 2. Suppose H_n also has the property that for each index i, h_ii = a, h_{i, i+1} = b and h_i,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 H_n = a (det H_n-1) - bc (det H_n-2) for n = 3,4,...

(ii) Find det H₆.

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 H_n-1 and H_n-2

 

[HallsofIvy]

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

Expand H_n on the first row. Then H_n= a(cofactor_a)- b(cofactor_b). It should be easy to see that "cofactor_a" is just H_n-1, so that part is aH_n-1 while "cofactor_b" has first column consisting of "c 0 0 ...". Expand that on the first column and you have "bcH_n-2".

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