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.
Projection Example 1