Gram Schmidt Process Algorithm

So I have an assignment to write algorithm code for Gram Schmidt process...I have done it in Matlab ,but when I run the code with input argument rand(some number), it gives me a result with variable ans... and some numbers in matrix... I want to create an output variable which won't be ans ,and can anyone see the code to tell me if it's okay? Thanks in advance,I would appreciate ur help..

    % function Q = gramschmidt( A )
% FUNCTION computes orthonormal basis from independent vectors in A.
%
%   Q = gramschmidt( A );
%
% INPUT :
%   A   - a matrix with *INDEPENDENT* vectors in columns
%         (e.g. A = rand(3) will produce one)
%
% OUTPUT :
%   Q   - a matrix with orthonomal basis of A
%
%

% The vectors in A are independent BUT NOT YET orthonormal. Check A'*A. % If it is orthonormal, you should get strictly an identity matrix.

% number of vectors n = size( A, 2 );

% initialize output Q = zeros( n );

% turn every independent vector into a basis vector % (1) jth basis vector will be perpendicular to 1..j-1 previous found basis % (2) will be of length 1 (norm will be equal to 1) for j = 1 : n

    % pick the jth independent vector
    u = A( :, j );

    % special case for j = 1: we will not run the following for loop. Will
    % just normalize it and put as the first found basis. There are no
    % previous basis to make orthogonal to.

    % remove from raw "u" all components spanned on 1..j-1 bases, their
    % contributions will be removed
    % ==> this effectively makes jth independent vector orthogonal to all
    % previous bases found in the previous steps.
    % ==> enforcing orthogonality principle here. Not orthonormality yet.
    for i = 1 : j - 1
        u = u - proj( Q(:,i), A(:,j) );
    end

    % normalize it to length of 1 and store it
    Q(:,j) = u ./ norm( u );

end

end

% projects a vector "a" on a direction "e" function p = proj( e, a )

% project "a" onto "e": (e' * a) / (e' * e) is the length (!) of "a" on "e" % multiplication by "e" is necessary to output the resulting vector which is % (colinear with "e") p = (e' * a) / (e' * e) .* e;

end

end

Answer 1

[Sebastián Soto]

for i = 1 : j - 1

u = u - proj( Q(:,i), A(:,j) );

end

% normalize it to length of 1 and store it

Q(:,j) = u ./ norm( u );

Answer 2

[Yashwanth]

%ECE24063

t1 = 0:0.01:1;

t2 = 1:0.01:2;

t3 = 2:0.01:3;

t = [t1,t2,t3];

s1 = ones(1,length(t1));

s2 = ones(1,length(t2));

s3 = zeros(1,length(t3));

S_1 = [s1 s2 s3];

s4 = ones(1,length(t1));

s5 = -1*ones(1,length(t2));

s6 = zeros(1,length(t3));

S_2 = [s4 s5 s6];

s7 = ones(1,length(t1));

s8 = ones(1,length(t2));

s9 = ones(1,length(t3));

S_3 = [s7 s8 s9];

figure;

subplot(3,1,1)

plot(t,S_1)

title("S1 signal")

xlabel("Time")

ylabel("Amplitude")

subplot(3,1,2)

plot(t,S_2)

title("S2 signal")

xlabel("Time")

ylabel("Amplitude")

subplot(3,1,3)

plot(t,S_3)

title("S3 signal")

xlabel("Time")

ylabel("Amplitude")

e1 = sum(S_1.*S_1)*0.01;

ps1 = S_1/(sqrt(e1));

c21 = sum(S_2.*ps1)*0.01;

d2 = S_2 - (c21*ps1);

e2 = sum(d2.*d2)*0.01;

ps2 = d2/(sqrt(e2));

c31 = sum(S_3.*ps1)*0.01;

c32 = sum(S_3.*ps2)*0.01;

d3 = S_3 - ((c31*ps1)+(c32*ps2));

e3 = sum(d3.*d3)*0.01;

ps3 = d3/(sqrt(e3));

figure;

subplot(3,1,1)

plot(t,ps1)

title("1st basis function")

xlabel("Time")

ylabel("Amplitude")

subplot(3,1,2)

plot(t,ps2)

title("2nd basis function")

xlabel("Time")

ylabel("Amplitude")

subplot(3,1,3)

plot(t,ps3)

title("3rd basis function")

xlabel("Time")

ylabel("Amplitude")

c11 = sum(S_1.*ps1)*0.01;

c12 = sum(S_1.*ps2)*0.01;

c13 = sum(S_1.*ps3)*0.01;

S1 = [c11 c12 c13];

display(S1)

c22 = sum(S_2.*ps2)*0.01;

c23 = sum(S_2.*ps3)*0.01;

S2 = [c21 c22 c23];

display(S2)

c33 = sum(S_3.*ps3)*0.01;

S3 = [c31 c32 c33];

display(S3)