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 :

% 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

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