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hw3solutions_2

# hw3solutions_2 - Homework 1 Solutions Josh Hernandez 1 2.1...

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Homework 1 Solutions Josh Hernandez October 27, 2009 1 2.1 - Linear Transformations, Null Spaces, and Ranges For 3 and 6, prove that T is a linear transformation, find bases for both N ( T ) and R ( T ), compute the nullity and rank of T , and verify the dimension theorem. Finally, determine whether T is one-to-one or onto: 3 T : R 2 R 3 defined by T ( a 1 , a 2 ) = ( a 1 + a 2 , 0 , 2 a 1 - a 2 ). Solution: Linearity: T ( ( a 1 , a 2 ) + c ( b 1 , b 2 ) ) = T ( ( a 1 + cb 1 , a 2 + cb 2 ) ) = ( ( a 1 + cb 1 ) + ( a 2 + cb 2 ) , 0 , 2( a 1 + cb 1 ) - ( a 2 + cb 2 ) ) = ( ( a 1 + a 2 ) + c ( b 1 + b 2 ) , 0 , (2 a 1 - a 2 ) + c (2 b 1 - b 2 ) ) = ( a 1 + a 2 , 0 , 2 a 1 - a 2 ) + c ( b 1 + b 2 , 0 , 2 b 1 - b 2 ) = T ( a 1 , a 2 ) + c T ( b 1 , b 2 ) . As for the nullspace, N ( T ) = { ( a 1 , a 2 ) : a 1 + a 2 = 2 a 1 - a 2 = 0 } The linear system a 1 + a 2 = 0 2 a 1 + - a 2 = 0 has only the trivial solution. Therefore, the nullspace of T is { 0 } . The range of T is span { (1 , 0 , 2) , (1 , 0 , -1) } = span { (1 , 0 , 0) , (0 , 0 , 1) } ; each of these sets is also a basis of R ( T ). Thus rank( T ) = 2. This verifies the rank-nullity theorem: 2 = dim( V ) = nullity( T ) + rank( T ) = 0 + 2 . This function is one-to-one, since its nullity is zero. It is not onto, since the second coordinate of any vector in the range must be zero. 6 T : M n × n ( F ) F defined by T ( A ) = tr( A ). Solution: Linearity: T ( A + cB ) = n X i =1 ( A + cB ) ii = n X i =1 ( A ii + cB ii ) = n X i =1 A ii + c n X i =1 B ii = T ( A ) + c T ( B ) .

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