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2.A. Proof of theorem 2.2.1
49
(ii) Assume r < n ⤠r + s. Let M2 = (mr +1 , . . . , mr +s ). Deï¬ne Z = XA,
where the (r + s) Ã (r + s) matrix
1
â¦â 2
A=
0
0
Is
is of rank r + s. We partition Z as follows:
1
Z = (Y â¦â 2 , M2 ) = (Z1 , M2 ),
1
1
where Z1 = Y â¦â 2 â¼ N nÃr (M1 â¦â 2 , In â Ir ). Let Z1 = (Z11 , . . . , Z1r ),
then by construction Z11 , . . . , Z1r are independently distributed with
Z1j ⼠N n (λj , In ),
j = 1, . . . , r
and
1
M1 â¦â 2 = (λ1 , . . . , λn ).
Subsequently, consider the n à n submatrix ZÌ of Z deï¬ned by ZÌ =
(Z1 , mr +1 , . . . , mn ), then
P rank(ZÌ) < n
â¤
r
P Z1j â lh{Z11 , . . . , Z1,jâ1 , Z1,j+1 , . . . , Z1r , mr +1 , . . . , mn }
j=1
=
0.
This implies that
P (rank(X) = n) = P (rank(Z) = n) ⥠P rank(ZÌ) = n = 1.
(iii) Assume that n > r + s. Let M2 = (m1 , . . . , mn ). Select mi1 , . . . , mir +s
such that the (r + s) à s matrix M3 deï¬ned by M3 = (mi1 , . . . , mir +s )
is of rank s. Let Y = (Y1 , . . . , Yn ) and deï¬ne Z = (Yi1 , . . . , Yir +s ). Now
we consider the (r + s) Ã (r + s) random matrix (Z, M3 ). According to
part (ii) we have
P rank(Z, M3 ) = r + s = 1.
Since X is obtained from (Z, M3 ) by adding rows we now may conclude
that rank(X) = r + s with probability one.
As an immediate extension we have