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Re: Iterative Procedure Using Vectors
Posted:
Apr 26, 2013 3:26 AM


On 4/25/2013 12:56 PM, Laura wrote: > I have been told to carry out the following iterative procedure: > a) Set a vector v to be the vector [1 ; 0] > b) Find a new vector u = Av (where A is a 2 x 2 real matrix) > c) Find the magnitude of this new vector u, and call this magnitude c > d) Divide u by c > e) Set v to be this unit vector, then repeat steps b) to e) a number of times. > > So far I have tried: > v(1) = [1 ; 0] > for n = 1:100 > u(n) = A*v(n+1) > c(n) = norm(u(n)) > v(n+1) = u(n)/c(n) > end > > However, it doesn't seem to like the v(1) part and keeps saying submatrix error, >probably because it thinks that I am referring to the first element in the vector >instead of saying v = [1 ; 0] is the initial condition. Does anyone know how I can do this? > Also I'm not sure if the rest of it works ... >
There is nothing in the HW instructions that says you need to keep track of all the v's. So why not just
 v = [1;0]; A = rand(2); for n = 1:10 u = A*v; v = u/norm(u) end 
If you need to save all the v's, then try
 N = 10; v = zeros(2,N); v(:,1) = [1;0]; A = rand(2);
for n = 2:N u = A*v(:,n1); v(:,n) = u/norm(u); end 
EDU>> v'
ans =
1.000000000000000 0 0.942479907869958 0.334262805680553 0.941008227921574 0.338383680138241 0.940991359516029 0.338430585668871 0.940991167526323 0.338431119487329 0.940991165341345 0.338431125562555 0.940991165316478 0.338431125631695 0.940991165316195 0.338431125632482 0.940991165316192 0.338431125632491 0.940991165316192 0.338431125632491
that is a good convergence. You only need 34 iterations really.
Nasser



