ans1 - Math 423, Answers for homework 1 1.2.1. (a) True, by...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 423, Answers for homework 1 1.2.1. (a) True, by definition. (b) False (homework). (c) False: take x = 0 . Comment: true if x 6 = 0 . (d) False: take a = 0 . Comment: true if a 6 = 0 . (e) True. (f) False: m rows, n columns. (g) False. (h) False: highest terms may cancel. (i) True. (j) True. (k) True. 1.2.2. 0 0 0 0 0 0 0 0 0 0 0 0 . 1.2.7. We observe that f (0) = 1 , f (1) = 3 , g (0) = 1 , g (1) = 3 , h (0) = 2 , h (1) = 6 . 1.2.9. Corollary 1. The vector described in (VS 3) is unique. Proof. We need to show that if there is another element z V such that for all x V , x + z = x , then z = 0 . Assume that z has the above property. Consider z + 0 . On the one hand, it is equal to z by (VS 3). On the other hand, it is equal to 0 + z by (VS 1), which is equal to by the defining property of z . We conclude that z = z + 0 = 0 . Another proof: since x + 0 = x = x + z , by (VS 1) we conclude that 0 + x = z + x , and by cancellation it follows that z = 0 . Corollary 2. The vector y described in (VS 4) is unique. Proof. We fix x V . The vector y V is defined by the property that x + y = 0 . Suppose there is another vector z V , also with the property x + z = 0 . Consider....
View Full Document

This note was uploaded on 01/19/2011 for the course MATH 121a taught by Professor Staff during the Winter '08 term at UC Irvine.

Page1 / 3

ans1 - Math 423, Answers for homework 1 1.2.1. (a) True, by...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online