Page 124 - Applied Numerical Methods Using MATLAB

PROBLEMS 113
(a) Consider the problem of solving

 
x 1
123 6
A 1 x =  x 2  = = b 1 (P2.8.3)
246 12
x 3
Since this belongs to the underdetermined case (M = 2 < 3 = N), it
seems that we can use Eq. (2.1.7) to find the minimum-norm solution.
(i) Type the following statements into the MATLAB command window.
>>A1 = [1 2 3; 2 4 6]; b1 = [6;12]; x = A1’*(A1*A1’)^-1*b1 %Eq. (2.1.7)
What is the result? Explain why it is so and support your answer by
typing
>>r = rank(A1)

(ii) Type the following statements into the MATLAB command window
to see the SVD-based minimum-norm solution. What is the value of
−1
ˆ ˆ −1 ˆ T
x = A b 1 = V S U b 1 and ||A 1 x − b 1 ||?
ˆ
1
[U,S,V] = svd(A1); %(P2.8.1)
u = U(:,1:r); v = V(:,1:r); s = S(1:r,1:r);
AIp = v*diag(1./diag(s))*u’; %faked pseudo-inverse (P2.8.2)
x = AIp*b1 %minimum-norm solution for singular underdetermined
err = norm(A1*x - b1) %residual error

(iii) To see that the norm of this solution is less than that of any other
solution which can be obtained by adding any vector in the null space
of the coefficient matrix A 1 , type the following statements into the
MATLAB command window. What is implied by the result?

nullA = null(A1); normx = norm(x);
for n = 1:1000
if norm(x + nullA*(rand(size(nullA,2),1)-0.5)) < normx
disp(’What the hell smaller-norm sol - not minimum norm’);
end
end
(b) For the problem

 
x 1
12 3 6
A 2 x =  x 2  = = b 2 (P2.8.4)
23 4 9
x 3
compare the minimum-norm solution based on SVD and that obtained
by Eq. (2.1.7).