hw4sol

# hw4sol - 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 4 Problems 3.3, 3.4, 3.5, 3.7 and 3.9 from Axler. Solutions. 3.3 Let U ⊂ V be a subspace and S : U → W be a linear map. Pick a basis ( u 1 ,...,u ‘ ) of U and extend it to a basis of V , ( u 1 ,...,u ‘ ,v ‘ +1 ,...,v n ) . Define a map T : V → W by T ( u i ) = S ( u i ) and T ( v j ) = W and extending by linearity. If u ∈ U then write u = ∑ i c i u i and see that T ( u ) = ∑ i c i S ( u i ) = S ( u ), as we needed to show. 3.4 If T is the zero map then null( T ) = V and the question doesn’t really make sense. Assume that T is not the zero map and pick some element u that is not mapped to zero. The intersection null( T ) with { au : a ∈ F } is zero, since u / ∈ null( T ). We apply the rank nullity theorem to see that dim(null( T )) = dim( V )- rank( T ). Any non-zero linear map V → F is surjective, since any non- zero element of F is a scalar multiple of another. Thus dim(null( T )) = dim( V )- 1. This means V = null( T ) + { au : a ∈ F } , since the sum is at least n dimensional. 3.5 Suppose that ( v 1 ,...,v n ) is linearly independent. If ( T ( v 1 ) ,...,T ( v n )) were linear dependent there would be scalars a i , not all zero, such that ( * ) X i a i T ( v i ) = = ⇒ T ( X i a i v i ) = That is, ∑ i a i v i is in the null space of T . But if T is injective, the null space is zero. This means ∑ i a i v i...
View Full Document

## This note was uploaded on 11/13/2011 for the course MATH 67 taught by Professor Schilling during the Winter '07 term at UC Davis.

### Page1 / 4

hw4sol - 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