This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS #1. Page 242; Exercise 2. Which of the following sets of vectors are bases for R 3 ? (a) 1 2 , 1 1 (b) 1 1 1 , 2 3 4 , 4 1 1 , 1 1 (c) 3 2 2 ,  1 2 1 , 1 (d) 1 , 2 1 , 3 4 1 , 1 Solution: (a) Not a basis. The vector space R 3 has dimension 3. According to Corollary 4.5, any subset of 2 < 3 vectors cannot span R 3 . Hence this subset cannot be a basis. (b) Not a basis. According to Corollary 4.4, any subset of 4 > 3 vectors must be linearly dependent. Hence this subset cannot be a basis. (c) A basis. Since this subset has 3 vectors, if we show that it is linearly independent then we may conclude it is a basis by Theorem 4.12. Indeed, consider the equation a 1 3 2 2 + a 2  1 2 1 + a 3 1 = = ⇒ 3 a 1 a 2 = 0 2 a 1 + 2 a 2 + a 3 = 0 2 a 1 + a 2 = 0 We find the following augmented matrix and its reduced row echelon form: 3 1 0 2 2 1 2 1 0 = ⇒ 1 0 0 0 1 0 0 0 1 = ⇒ a 1 = 0 a 2 = 0 a 3 = 0 Hence the only solution is a 1 = a 2 = a 3 = 0. This shows that the subset is linearly independent, and so it must be a basis for R 3 . (d) Not a basis. According to Corollary 4.4, any subset of 4 > 3 vectors must be linearly dependent. Hence this subset cannot be a basis. #2. Page 242; Exercise 3. Which of the following sets of vectors are bases for R 4 ? (a) 1 0 0 1 , 0 1 0 0 , 1 1 1 1 , 0 1 1 1 (b) 1 1 0 2 , 3 1 2 1 , 1 0 0 1 (c) 2 4 6 4 , 0 1 2 0 , 1 2 3 2 , 3 2 5 6 , 2 1 0 4 1 2 MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS (d) 0 0 1 1 , 1 1 1 2 , 1 1 0 0 , 2 1 2 1 Solution: (a) A basis. The vector space R 4 has dimension 4. Since this subset has 4 vectors, if we show that it is linearly independent then we may conclude it is a basis by Theorem 4.12. Indeed, consider the equation a 1 1 0 0 1 + a 2 0 1 0 0 + a 3 1 1 1 1 + a 4 0 1 1 1 = 0 0 0 0 We may express this as the linear system a 1 + a 3 = 0 a 2 + a 3 + a 4 = 0 a 3 + a 4 = 0 a 1 + a 3 + a 4 = 0 = ⇒ 1 0 1 0 0 1 1 1 0 0 1 1 1 0 1 1 = ⇒ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 Hence the only solution is a 1 = a 2 = a 3 = a 4 = 0. This shows that the subset is linearly independent, and so it must be a basis for R 4 . (b) Not a basis. According to Corollary 4.5, any subset of 3 < 4 vectors cannot span R 4 . Hence this subset cannot be a basis....
View
Full Document
 Spring '10
 ...
 Linear Algebra, Vector Space, basis, A3 A3

Click to edit the document details