Page 256 - MATLAB an introduction with applications

Numerical Methods ——— 241

Q(1:k) = zeros(1,k);
Q(k+1:n) = V(k+1:n)–c*W(k+1:n);
A(k+2:n,k) = zeros(n–k–1,1);
A(k,k+2:n) = zeros(1,n–k–1);
A(k+1,k) = –s;
A(k,k+1) = –s;
A(k+1:n,k+1:n) = A(k+1:n,k+1:n) ...
–2*W(k+1:n)’*Q(k+1:n)–2*Q(k+1:n)’*W(k+1:n);
end
T = A;
fprintf(‘Matrix in tridiagonal form is\n’);
disp(T);
The output of the program is given below:
Matrix in tridiagonal form is
4.0000 –3.7417 0 0
–3.7417 6.5714 2.7180 0
0 2.7180 3.0529 1.2403
0 0 1.2403 2.3757

Example E4.19: Use the Sturn sequence property to find the interval of the smallest eigenvalue of

 2 − 1 0 0 
 − 1 2 − 2 0 
A =  
 0 − 2 2 − 1
 
  0 0 − 1 2  

Solution: The sequence {f (a)} and {f (b) can be used to determine the number of roots of f (λ), which are
k
k
n
contained in [a, b].
The sequence {f , f ,..., fn} forms a Sturn sequence of polynomials; and such sequences have special
0
1
properties. Given a point b, calculate
{ ( ), ( ),...,fb f b f n ( )}
b
0
1
0
()
()
and observe the signs of these quantities. If some f λ= , then choose the sign of f λ to be opposite
j
j
to that of f j− 1 () λ . It can be shown that
f λ= ⇒ f j− 1 () λ≠ 0
()
0
j
λ
Having obtain a sequence of signs from flet s () denote the number of agreements of sign between
consecutive members of the sign sequence.