hmw3 - Solutions to Homework 3. Math 110, Fall 2006. Prob...

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Solutions to Homework 3. Math 110, Fall 2006. Prob 2.1.10. By the linearity of T , we use the fact - (1 , 0) + 3(1 , 1) = (2 , 3) to obtain T (2 , 3) = - T (1 , 0) + 3 T (1 , 1) = - (1 , 4) + 3(2 , 5) = (5 , 11) . The map T is 1 - 1 as we see that T ( a (1 , 0)+ b (1 , 1)) = a (1 , 4)+ b (2 , 5) implies a = b = 0. Thus N ( T ) = { 0 } , so T is 1 - 1. Prob 2.1.15. The map T is linear, since indefinite integration is linear: Z x 0 ( αf ( t ) + βg ( t )) dt = α Z x 0 f ( t ) dt + β Z x 0 g ( t ) dt. To see that T is onto, suppose Tf is the identically zero polynomial. By differentiating Tf , we obtain f = 0. Hence the kernel of T is trivial. Finally, notice that all polynomials of the form Tf are zero at x = 0. So we cannot get any polynomial with nonzero constant term by applying T , i.e., T is not onto. Prob 2.1.28. T maps 0 to 0 by linearity and maps V to V by definition, these two subspaces are trivially T -invariant. If x N ( T ), then Tx = 0 N ( T ), so N ( T ) is T -invariant. If x
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This note was uploaded on 10/02/2009 for the course MATH 54554 taught by Professor Holtz during the Fall '06 term at University of California, Berkeley.

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hmw3 - Solutions to Homework 3. Math 110, Fall 2006. Prob...

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