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Jacobi Matrix

  1. Jul 26, 2009 #1
    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. jcsd
  3. Jul 26, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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
    [tex]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|[/tex]
    [tex]= 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[/tex]
     
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