Rudin Solutions Chapter 9 73931

MATH 205 SECTION 41 HW #1 RUDIN SOLUTIONS

Rudin 9.1 Let B =< v1 , v2 , · · ·, vk > be a basis for S. Take w, z∈SpanS. Then w = c1 v1 +· · ·+ck vk and z = d1 v1 +· · ·+dk vk for ci , di ∈R, where i = 1, 2, · · ·, k. So w+z = (c1 +d1 )v1 +· · ·+(ck +dk )vk , which is a linear combination of B since (ci + di )∈R for i = 1, 2, · · ·, k. Therefore, (w + z)∈SpanS. Also, ∀q∈R, we have that qw = qc1 v1 + · · · + qck vk , which implies that qw∈SpanS since qci ∈R for i = 1, 2, · · ·, k. Therefore, SpanS is a vector space. Rudin 9.2 Let A and B be linear transformations and take x, y∈X. Then BA(x + y) = B(A(x) + A(y)) = BA(x) + BA(y), and for c∈R, BA(cx) = B(cA(x)) = cBA(x), which shows that BA is linear. Now consider A−1 . Notice that A∈L(X, Y ), so A−1 : Y →X. Take x, y∈Y , then A−1 (A(x + y) = x + y = A−1 (A(x)) + A−1 (A(y)), and for c∈R, A−1 (A(cx)) = cx = cA−1 (A(x)), showing that A−1 is linear. Lastly, we want to show that A−1 is invertible. By the existence of A−1 we know that A is invertible, so A is one to one, meaning that A(x) = A(y) if and only if x = y. So A(x) = A(y) if and only if A−1 (A(x)) = A−1 (A(y)), showing that A−1 is one to one. Also, A is onto, meaning that R(A) = Y . Since A : X→Y , we have that A−1 (Y ) = X, meaning A−1 is onto. Therefore, A−1 is invertible. Rudin 9.3 Take A∈L(X, Y ). Suppose there exist x, y∈X such that A(x) = A(y). Then A(x) − A(y) = A(x − y) = 0, which implies that x = y since A(z) = 0 only when z = 0. Therefore, A is one to one.

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