We shall define the vector Zi yi / i and the matrix Fij fj(xi) / i.
Note that H = FT . F by Eq. (27),
(33) |
Then
(34) |
where the unstarred is used for 0.
using Eq. (34). The second term on the right is zero because of Eq. (33).
(34) |
Note that
If qi is an eigenvalue of Q, it must be equal qi2, an eigenvalue of Q2. Thus qi = 0 or 1. The trace of Q is
Since the trace of a matrix is invariant under a unitary transformation, the trace always equals the sum of the eigenvalues of the matrix. Therefore M of the eigenvalues of Q are one, and (p - M) are zero. Let U be the unitary matrix which diagonalizes Q (and also (1 - Q)). According to Eq. (35),
Thus
where S* is the square of the radius vector in (p - M)-dimensional space. By definition (see Section 16) this is the 2 distribution with (p - M) degrees of freedom.