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.