Matlab and Octave have a function orth() which will compute an orthonormal basis for a space given any set of vectors which span the space. In Matlab, e.g., we have the following help info:
>> help orth ORTH Orthogonalization. Q = orth(A) is an orthonormal basis for the range of A. Q'*Q = I, the columns of Q span the same space as the columns of A and the number of columns of Q is the rank of A. See also QR, NULL.
Below is an example of using orth() to orthonormalize a linearly independent basis set for :
% Demonstration of the orth() function. v1 = [1; 2; 3]; % our first basis vector (a column vector) v2 = [1; -2; 3]; % a second, linearly independent vector v1' * v2 % show that v1 is not orthogonal to v2 ans = 6 V = [v1,v2] % Each column of V is one of our vectors V = 1 1 2 -2 3 3 W = orth(V) % Find an orthonormal basis for the same space W = 0.2673 0.1690 0.5345 -0.8452 0.8018 0.5071 w1 = W(:,1) % Break out the returned vectors w1 = 0.2673 0.5345 0.8018 w2 = W(:,2) w2 = 0.1690 -0.8452 0.5071 w1' * w2 % Check that w1 is orthogonal to w2 ans = 2.5723e-17 w1' * w1 % Also check that the new vectors are unit length ans = 1 w2' * w2 ans = 1 W' * W % faster way to do the above checks ans = 1 0 0 1 % Construct some vector x in the space spanned by v1 and v2: x = 2 * v1 - 3 * v2 x = -1 10 -3 % Show that x is also some linear combination of w1 and w2: c1 = x' * w1 % Coefficient of projection of x onto w1 c1 = 2.6726 c2 = x' * w2 % Coefficient of projection of x onto w2 c2 = -10.1419 xw = c1 * w1 + c2 * w2 % Can we make x using w1 and w2? xw = -1 10 -3 error = x - xw error = 1.0e-14 * 0.1332 0 0 norm(error) % typical way to summarize a vector error ans = 1.3323e-15 % It works! (to working precision, of course)
% Construct a vector x NOT in the space spanned by v1 and v2: y = [1; 0; 0]; % Almost anything we guess in 3D will work % Try to express y as a linear combination of w1 and w2: c1 = y' * w1; % Coefficient of projection of y onto w1 c2 = y' * w2; % Coefficient of projection of y onto w2 yw = c1 * w1 + c2 * w2 % Can we make y using w1 and w2?
yw = 0.1 0.0 0.3 yerror = y - yw yerror = 0.9 0.0 -0.3 norm(yerror) ans = 0.9487
While the error is not zero, it is the smallest possible error in the least squares sense. That is, yw is the optimal least-squares approximation to y in the space spanned by v1 and v2 (w1 and w2). In other words, norm(yerror) is less than or equal to norm(y-yw2) for any other vector yw2 made using a linear combination of v1 and v2. In yet other words, we obtain the optimal least squares approximation of y (which lives in 3D) in some subspace (a 2D subspace of 3D spanned by the columns of matrix W) by projecting y orthogonally onto the subspace to get yw as above.
An important property of the optimal least-squares approximation is that the approximation error is orthogonal to the the subspace in which the approximation lies. Let's verify this:
W' * yerror % must be zero to working precision ans = 1.0e-16 * -0.2574 -0.0119