Exercise 10.6 Show that V

1(I − M)V1 is invertible.

Answer: The columns of V1 form a basis, so 0 = V1b iff b = 0. Also

(I − M)V1b = 0 iff V1b ∈ C(X), but V1b ∈ C(X) iff b = 0 by choice of V1.

Thus, (I − M)V1b = 0 iff b = 0; hence (I − M)V1 has full column rank and V

1(I − M)V1 is invertible.