Math121sol2

# Math121sol2 - Math 121 Linear Algebra and Applications...

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

Math 121: Linear Algebra and Applications Prof. Lydia Bieri Solution Set 2 Posted: Fri. Oct. 17, 2007 Written by: Luca Candelori Exercise 1 (1.4/12) . Recall that for ’if and only if’ statements we need to prove both directions: assuming the statement to the left prove the one to the right, and viceversa. In our case, suppose that W is a subspace of V . For any set S , it is always true that S span( S ), so obviously W span( W ). Take now a basis for W , call it w 1 ,...,w k . Then a typical element of span( W ) is given by w = k X i =0 a i w i but this is in W , since W is closed under addition and scalar multiplication. Hence W = span( W ). Conversely, suppose that span( W ) = W . We know that 0 span( W ), since 0 · w is in span( W ). Therefore 0 W , since they are equal. Let now v,w W . Then v + w span( W ) by deﬁnition, and so is c · v for any c and v . Therefore W is a subspace. Exercise 2. (a) (1.5/4) Suppose the vectors e 1 ,...,e n are linearly dependent. Then we can ﬁnd scalars a 1 ,...,a n , not all zero, such that a 1 1 0 . . . 0 0 + a 2 0 1 . . . 0 0 + ... + a n 0 0 . . . 0 1 = 0 0 . . . 0 0 but now, looking at each component, we get that a 1 = a 2 = ... = a n = 0, which is a contradiction. Therefore e 1 ,...,e n are linearly independent. (b) (1.5/5) Again, we apply the same method. Suppose there exist scalars a 1 ,...,a n , not all zero, such that f ( x ) = a 1 + a 2 x + ... + a n x n = 0 Then the polynomial f is of degree k , for some 0 k n , and so it should have at most k roots. But this polynomials is zero everywhere, which implies that a 1 = a 2 = ... = a n = 0. Exercise 3. 1

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

View Full Document
(i) Suppose S = { s 1 ,s 2 ,...,s k } is a generating set for V with k < n elements. Since S generates V , we should have span( S ) = V . But dim(span( S )) k < n , which is a contradiction. Therefore any generating set S must have k n . If k = n , and S is not a basis for V , that means that some element of S , say s k , is linearly dependent on s 1 ,...,s k . We can throw away s k to obtain a new set S 1 = S - { s k } which still spans V , i.e. span( S 1 ) = V . But dim(span( S 1 )) k - 1 = n - 1, which is a contradiction. Therefore
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/23/2010 for the course MATH Math 121 taught by Professor Bieri during the Fall '07 term at Harvard.

### Page1 / 5

Math121sol2 - Math 121 Linear Algebra and Applications...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online