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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 $ 1.0$ and the next representable number in double-precision floating point.


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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8
Copyright © 2024-02-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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