Projection Example 2

Let X and PX be defined as Example 1, but now let

>> y = [1;-1;1]
y =

   1
  -1
   1

>> yX = PX * y
yX =

   1.33333
  -0.66667
   0.66667

>> yX' * (y-yX)
ans = -7.0316e-16

>> eps
ans =  2.2204e-16

In the last step above, we verified that the projection yX is orthogonal to the ``projection error'' y-yX, at least to machine precision. The eps variable holds ``machine epsilon'' which is the numerical distance between and the next representable number in double-precision floating point.

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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.