#### 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.

**Previous Section:**

Projection Example 1