This preview shows pages 1–3. Sign up to view the full content.
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
 ...

Click to edit the document details