Page 173 - Matrices theory and applications

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Lemma 9.4.1 Let µ be a nonzero complex number and C a tridiagonal
matrix, of diagonal C 0 , of upper triangular part C + and lower triangular
part C − .Then

1
C − + µC + .
det C =det C 0 +
µ
Proof 9. Iterative Methods for Linear Problems
It is enough to observe that the matrix C is conjugate to
1
C 0 + C − + µC + ,
µ
through the linear transformation matrix
 
µ
µ 2 0
 
 
.
 . 
Q µ =  .  .
 
.
 . 
0 .
 
µ n
Let us apply the lemma to the computation of the characteristic
polynomial P ω of L ω .We have
(det D)P ω (λ) = det((D − ωE)(λI n −L ω ))
= det((ω + λ − 1)D − ωF − λωE)
λω

= det (ω + λ − 1)D − µωF − E ,
µ
for every nonzero µ.Let us choose for µ any square root of λ.We thenhave
2
2
(det D)P ω (µ ) = det((ω + µ − 1)D − µω(E + F))
2
= (det D)det((ω + µ − 1)I n − µωJ).
Finally, we have the following lemma.
Lemma 9.4.2 If A is tridiagonal and D invertible, then
2
µ + ω − 1
2 n
P ω (µ )= (µω) P J ,
µω
where P J is the characteristic polynomial of the Jacobi matrix J.
Let us begin with the analysis of a simple case, that of the Gauss–Seidel
method, for which G = L 1 .
Proposition 9.4.1 If A is tridiagonal and D invertible, then:
2
n
1. P G (X )= X P J (X),where P G is the characteristic polynomial of
the Gauss–Seidel matrix G,