Math416_Final_Practice_sol - Math 416 - Abstract Linear...

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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 finite-dimensional 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....
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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.

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Math416_Final_Practice_sol - Math 416 - Abstract Linear...

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