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Unformatted text preview: Math 311, Homework 7 partial answers and solutions 3.4.5. f : R 3 → R 2 is a linear function, and has a matrix representation f x y z = 1 4 3 2 5 4 x y z . So its image is the span Span 1 2 , 4 5 , 3 4 Since 1 = 3 4 2 1 2 and 1 = 1 2 + 3 4 4 5 , this span is in fact all of R 2 , so the function is onto. Finally, to find its null space we solve 1 4 3 2 5 4 x y z = . Using row reduction this is equivalent to 1 4 3 3 2 x y z = , whose solutions are x y z = 8 t 9 t 2 t 3 t = Span  1 2 3 . 3.4.6. Answer: R 3 , not a subspace, not linear. 3.4.9. The image consists of all functions g ( x ) in C (1) (∞ , ∞ ) with g (0) = 0 . Indeed, clearly any g = F ( u ) has to satisfy this condition. Conversely, take any g like this, and let u ( x ) = e x g ( x ) . Then u is in C ( ∞ , ∞ ) and F ( u )( x ) = g ( x ) . The function is linear. If F ( u )( x ) = 0 for all x , then F ( u ) ( x ) = 0 for all x . But this means that e x u ( x ) = 0 for all x , so that u ( x ) = 0 for all x , and the nullspace of F contains only the zero function. 1 3.4.13. The image is Span 2 2 , 4 1 1 , 1 1 = Span 4 1 1 , 1 1 . The nullspace consists of all the solutions of 2 4 1 0 1 0 2 1 1 x y z = , which is equivalent to 2 4 1 0 1 0 0 0 0...
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This note was uploaded on 04/29/2008 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.
 Spring '08
 Anshelvich
 Math

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