For loop help for Linear Algebra Class

I cant seem to get this algorithm code correct.

%HW3 #2
% Given:
%
% y_k denotes k index
%
% y_k=A*x_k
% r_k=norm(y_k) / norm(x_k)
% x_(k+1) = y_k / norm(y)
% 
% Create a vector with 0-50 indexes of k

clear all;
close all;
A = [-5,3,-1;0,4,1;0,0,-2];
x(1)=[1;1;1];
for k = 1:51
    y{k} = A.*x;
    r{k} = norm(y)/norm(x);
    x{k+1} = y/norm(y);

end

I receive the error: "In an assignment A(I) = B, the number of elements in B and I must bethe same."

Im not sure how to go about fixing the problem, because I do not fully understand how mat lab operates yet.

[Cedric]

In addition to Walter's answer, y in your loop is the full cell array. It is probably not what you want to use when you compute x{k+1}.

[Sean Murphy]

It seemed that I had to define each variable with the cell array index in order to obtain no errors.

[Sean Murphy]

Thanks guys!!

Fixed code:

%HW3 #2
% y_k=A*x_k
% r_k=||y_k||_2 / ||x_k||_2
% x_k+1 = y_k / ||y_k||_2
% r = norm(y0,2)/norm(x0,2)
% x_k1 = y0/norm(y0,2)
clear all;
close all;
A = [-5,3,-1;0,4,1;0,0,-2];
x{1}= [1;1;1];
for k = 1:51
    y{k} = A*x{k};
    r{k} = norm(y{k})/norm(x{k});
    x{k+1} = y{k}/norm(y{k});
end
q_x = [x{1,51}]
q_y = [y{1,51}]
q_r = [r{1,51}]
save x50.dat q_x -ascii
save y50.dat q_y -ascii
save r50.dat q_r -ascii

[Cedric]

You're most welcome! Note that x, y, r are 1D cell arrays, so x{1,51} is the same as x{51}, and the [ ] that you use in [x{1,51}] have no effect.

If you wanted to save all the x vectors though, you could concatenate all cells content of x (which is a cell array, so its elements are cells whose content are 3x1 numeric arrays) the following way: [x{:}], which would be a 3x52 array of vectors in column. This dimension should also make you realize that as you compute x{k+1}, the boundaries that you are defining might need some additional thoughts, especially if you wanted to use then simultaneously x and y (as [y{:}] is a 3x51 array)..

Answer 1

[Walter Roberson]

Instead of

x(1)=[1;1;1];

you would need

x{1}=[1;1;1];

to avoid that error.

However, then when you go to the A.*x you would fail in trying to multiply A by a cell array. Try

y{k} = A.*x{k}

[Cedric]

 y{k} = A*x{k}

[Walter Roberson]

True, it would be matrix multiplication desired in that case.

Note that your r{k} and x{k+1} assignments will need similar adjustments to use x{k} and y{k}