This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 416  Abstract Linear Algebra Fall 2011, section E1 Practice Final Name: Solutions • This is a (long) practice exam. The real exam will consist of 6 problems. • In the real exam, no calculators, electronic devices, books, or notes may be used. • Show your work. No credit for answers without justification. • Good luck! 1. /10 2. /15 3. /15 4. /10 5. /10 6. /10 7. /10 8. /10 9. /10 10. /15 11. /10 12. /10 Total: /135 1 Problem 1. (10 pts) Let V,W be finitedimensional vector spaces. Show that V is isomor phic to W if and only if V and W have the same dimension. ( ⇒ ) Let ϕ : V ’→ W be an isomorphism. Pick a basis { v 1 ,...v n } of V . Since ϕ is an isomor phism, it sends any basis to a basis, so that { ϕv 1 ,...,ϕv n } is a basis of W . Thus we have dim W = n = dim V . ( ⇐ ) Assume dim V = dim W = n . Pick a basis { v 1 ,...,v n } of V and a basis { w 1 ,...,w n } of W . Define the linear map ϕ : V → W v i 7→ w i . Since ϕ sends a basis of V to a basis of W , it is an isomorphism. 2 Problem 2a. (6 pts) Show that vectors v 1 ,...,v n ∈ V are linearly independent if and only if there is a linear map T : V → W such that Tv 1 ,...,Tv n are linearly independent. [The content is in the “if” direction.] ( ⇒ ) Assume v 1 ,...,v n are linearly independent. Take T = id: V → V , so that the vectors id( v 1 ) ,..., id( v n ) are linearly independent. ( ⇐ ) Assume there is a linear map T : V → W such that Tv 1 ,...,Tv n are linearly independent. We want to show v 1 ,...,v n are linearly independent. Consider the equation c 1 v 1 + ... + c n v n = 0 . Applying T to the equation yields T ( c 1 v 1 + ... + c n v n ) = T (0) = 0 c 1 ( Tv 1 ) + ... + c n ( Tv n ) = 0 which implies c 1 = 0 ,...,c n = 0 because the vectors Tv 1 ,...,Tv n are linearly independent. Therefore the vectors v 1 ,...,v n are linearly independent. Alternate (geometric) proof. Since Tv 1 ,...,Tv n are linearly independent, we have dim Span { Tv 1 ,...,Tv n } = n dim T Span { v 1 ,...,v n } = n and the inequalities n = dim T Span { v 1 ,...,v n } ≤ dim Span { v 1 ,...,v n } ≤ n are forced to be equalities, which yields dim Span { v 1 ,...,v n } = n. Therefore the vectors v 1 ,...,v n are linearly independent. 3 b. (4 pts) Let X := ( a,b ) ⊆ R be an open interval of the real line, and consider the vector space of smooth functions on X C ∞ ( X ) := { f : X → R  derivatives f ( n ) exist for all n ≥ 1 } . Let x ∈ X be some point in the interval X . Check that T x : C ∞ ( X ) → R 2 f 7→ f ( x ) f ( x ) is a linear map. For any f,g ∈ C ∞ ( X ) and scalars α,β ∈ R , we have T x ( αf + βg ) = ( αf + βg )( x ) ( αf + βg ) ( x ) = αf ( x ) + βg ( x ) αf ( x ) + βg ( x ) = α f ( x ) f ( x ) + β g ( x ) g ( x ) = αT x ( f ) + βT x ( g ) which means T x is linear....
View
Full
Document
This note was uploaded on 12/16/2011 for the course MATH 416 taught by Professor Frankland during the Fall '11 term at University of Illinois at Urbana–Champaign.
 Fall '11
 FRANKLAND
 Math, Linear Algebra, Algebra

Click to edit the document details