5.22 This exercise is regarding Example 5.8 (continued) in Section 5.3.2. For parts

(i) and (ii) you may use the following matrix identity in Appendix A.1.2: (D±

BA

−1B



)

−1 = D

−1 ∓ D

−1B(A ± B



D

−1B)

−1B



D

−1.

(i) Define H = δI + Z



Z. Show that B(γ ) = δ(X



X − X



ZH

−1Z



X)

−1X



(δI + ZZ



)

−1.

(ii) Furthermore, let Q = δI + Z



PZ. Show by continuing with (i) that B(γ ) = (X



X)

−1X

{I + ZQ

−1Z



(I − P)}(I − ZH

−1Z



).

(iii) Continuing with (ii), show that B(γ ) = (X



X)

−1X



(I − ZQ

−1Z



P).

(iv) Show that (X



X)

−1X

 = λ

−1/2 min (X



X).

(v) Write A = P



Z. Show that the positive eigenvalues of S = A(δI +

AA



)

−2A

 are λi(δ + λi )

−2, 1 ≤ i ≤ m, where λi, 1 ≤ i ≤ m, are the positive eigenvalues of AA

. [Hint: The positive eigenvalues of S are the same as those of U = (δI + AA



)

−1A



A(δI + AA



)

−1.] Use this result to show

(δI + A



A)

−1A

 ≤

Z

min λi > 0

√

λi

, δ>0, where λ1, . . . , λm are the eigenvalues of A



A.