We shall define the vector Zi yi / i and the matrix Fij fj(xi) / i.
Note that H = FT . F by Eq. (27),
where the unstarred is used for 0.
using Eq. (34). The second term on the right is zero because of Eq. (33).
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),
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.