hw3sol

# hw3sol - Winter 2011 • Math 67 • Linear Algebra...

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: Winter 2011 • Math 67 • Linear Algebra Homework 3 Problem 2.3 Suppose that ( v 1 ,...,v n ) is linearly independent and that ( v 1 + w,...,v n + w ) is linearl dependent. Then there are scalars a i , not all zero, such that ∑ n i =1 a i ( v i + w ) = 0. Expanding this and moving the terms with w ’s to the right side of the equation yields a 1 v 1 + ··· + a n v n =- ( a 1 + ··· + a n ) w The right side is not zero, since it was then we would see that ( v 1 ,...,v n ) was linearly dependent, which it is not. This means ( a 1 + ··· + a n ) 6 = 0 so we can divide it away and get- 1 a 1 + ··· + a n ( a 1 v 1 + ··· + a n v n ) = w. Thus, w is in the span of v 1 ,...,v n . Problem 2.4 The set of polynomials in P ( F ) of degree m is not a subspace of P ( F ), since it is not closed under additions. To wit, ( z + z 2 )- z 2 = z is the sum of two polynomials of degree 2. However, z is not a polynomial of degree 2....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

hw3sol - Winter 2011 • Math 67 • Linear Algebra...

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

View Full Document
Ask a homework question - tutors are online