I I understood you well.

Try to have a look at this:

The projection matrix P of any matrix A could be found by formula P = A*inv(A'*A)*A';

or on the file projectmat.m where link is here below:

http://web.mit.edu/18.06/www/Course-Info/Tcodes.html

I hope it will help a little.

thanks. Meanwhile I got this:

[R, pivcol] = rref(A);

C = A(:, pivcol);

P = C*inv(C'*C)*C' % projector matrix for CS of A

%proj = cross(CS(:,1),CS(:,2)) % vector orthogonal to CS vectors

z_1=zeros(4,1); % Init

for n = 1:size(CS(:,1),1)

z_1(n,:) = (dot(CS(n,1),CS(n,2)'))/norm(dot(CS(n,1),CS(n,2)'));

end

z_1

solved :)

my column / row space seems to be incorrect, so I tried to find it directly, but Matlab doesn't know what span(A) is...

% 1c+d

% Q = orth(A)

% CS = colspace(sym(A))

% RS = rowspace(A)

NS = null(A); % nullspace

NS = NS/norm(NS)

% NS_trans = null(A'); % transponated version

% NS_trans = NS_trans/norm(NS_trans)

% NS_easy = null(A,'r'); % easier version

% NS_easy = NS_easy/norm(NS_easy)

% NS_easy_trans = norm(null(A', 'r')); % transponated easier version

% NS_easy_trans = NS_easy_trans/norm(NS_easy_trans) % equal to left null space

LNS = span(conj(A')) % left null space

% [B, pivot] = rref(A); % reduced row epsilon form

% r = length(pivot);

%

% CS = A(:,pivot)/norm(A(:,pivot)) % matrix of pivot columns of A

%

% rs = [B(1,:);B(2,:);B(3,:)]; % basis for row space of A

% RS = B(1:r,:)/norm(B(1:r,:)); % "cleared" row space

CS = span(A);

CS = CS/norm(CS)

RS = span(conj(A'));

RS = RS/norm(RS)

Maybe have a look at this:

http://www.mathworks.com/matlabcentral/newsreader/view_thread/270525

Unfortunately this only checks if the vector v is a subspace of matrix A. But I don't have one to check, I need to find it. Only matrix A is given.

The Formula

P=A*inv(A'*A)*A' only works if the rank of A is equal to the numbers of columns of A.

Here are two alternative ways to compute the projector into the column space of A that work fine independently of the rank of A:

1.)

An SVD of A results in A=U*S*V'

Here S is diagonal, and S# ist something like its inverse, whose components are defined as

S#ij =0 for i ~= j

S#ii = 1/Sii if Sii ~= 0

S#ii = 0 if Sii = 0

P = U * S# * S * U'

2.)

Ao = orth(A) ist an orthonormal Matrix that spans the column space of A than

P = Ao*Ao'

Hi,

If A is a (m x n), then the projection of B (m x p) can be calculated as

P = A * ( A \ B )