Orthogonal Basis Computation

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 $N=3$:

% 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 $W$ (a 2D subspace of 3D spanned by the columns of matrix W) by projecting y orthogonally onto the subspace $W$ 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