practiveprobssol

# practiveprobssol - SOLUTIONS FOR PRACTICE PROBLEMS Section...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTIONS FOR PRACTICE PROBLEMS Section 1.1 3. (c) The plane in question is the set of all points of the form (- 8 , 2 , 0) + t ( (- 8 , 2 , 0)- (1 , 3 , 0) ) + s ( (- 8 , 2 , 0)- (6 ,- 5 , 0) ) = (- 8- 9 t- 14 s, 2- t + 7 s, 0) , with t, s R . Section 1.2 17. No , V is not a vector space. For instance, the condition (VS8) (one of the distributive laws) fails. To illustrate this, consider the scalars a = b = 1, and the vector x = (1 , 0). Then ( a + b ) x = 2 x = (1 , 0), yet ax + bx = x + x = (2 , 0). Section 1.3 10. First show that W 1 = { ( a 1 , . . . , a n ) F n : a 1 + . . . + a n = 0 } is a subspace of F n . We have to prove that W 1 is closed under addition and scalar multiplication. More precisely, suppose u = ( u 1 , . . . , u n ) and v = ( v 1 , . . . , v n ) are elements of W 1 , and c is a scalar. We have to show that u + v and cu also belong to W 1 . We have u + v = ( u 1 + v 1 , . . . , u n + v n ), and then ( u 1 + v 1 ) + . . . + ( u n + v n ) = ( u 1 + . . . + u n ) + ( v 1 + . . . + v n ) = 0 + 0 = 0 , hence u + v W 1 . Similarly, u = ( cu 1 , . . . , cu n ), and the equality cu 1 + . . . + cu n = c ( u 1 + . . . + u n ) = c 0 = 0 , which shows cu W 1 ....
View Full Document

## This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

### Page1 / 4

practiveprobssol - SOLUTIONS FOR PRACTICE PROBLEMS Section...

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

View Full Document
Ask a homework question - tutors are online