=+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.