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2. Two-Point Boundary Value Problems
56
This matrix is obviously symmetric, and one easily computes the eigenval-
ues
√
λ =3 ± 2 2,
which are both positive. Hence A is positive definite. However, we observe
that A is not diagonal dominant. Therefore, Proposition 2.1 is not sufficient
to guarantee that Algorithm 2.1 will work for all positive definite systems.
But, as mentioned above, we shall prove that all symmetric tridiagonal
matrices that are positive definite can be handled by the method.
Proposition 2.4 Consider a tridiagonal system of the form (2.19) and
assume that the corresponding coefficient matrix (2.18) is symmetric and
positive definite. Then the system has a unique solution that can be com-
puted by Algorithm 2.1.
Proof: We claim that δ k > 0 for k =1, 2,... ,n. Assume on the contrary
that δ 1 ,δ 2 ,... ,δ k−1 > 0 and that δ k ≤ 0 for some index k,1 ≤ k ≤ n.We
will show that this assumption leads to a contradiction.
n
Define the vector v ∈ R by
v k = 1 and v k+1 = v k+2 = ··· = v n =0
and
γ j
v j = − v j+1 for j = k − 1,k − 2,... , 1.
δ j
This vector v satisfies the system (2.21) with
c 1 = c 2 = ··· = c k−1 =0,c k = δ k ≤ 0,b k+1 = β k+1,b k+2 = ··· = b n =0.
However, since (2.19) and (2.21) are equivalent, we obtain by (2.22) that
Av = b, where

b 1 = b 2 = ··· = b k−1 = 0 and b k = c k = δ k ≤ 0.
Since A is positive definite and v k = 1, we know that

T
v Av > 0.
On the other hand, from the properties of the vectors v and b above, we
have
n

T
T
v Av = v b = v j b j = b k ≤ 0.
j=1
This is the desired contradiction.