235FA10-SAMPLEM2-sol - MTH 235 : Linear Algebra Sample...

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Unformatted text preview: MTH 235 : Linear Algebra Sample problems - Midterm 2 1. Prove that T is a linear transformation, and find bases for the kernel (null space) and the range of T . Is T one-to-one? Is T onto? (a) T : R 3 R 2 defined by T ( a 1 ,a 2 ,a 3 ) = ( a 1- a 2 , 2 a 3 ). Solution. Let = ( a 1 ,a 2 ,a 3 ) and = ( b 1 ,b 2 ,b 3 ) be two vectors in R 3 and c R any scalar. T ( c + ) = T ( ca 1 + b 1 ,ca 2 + b 2 ,ca 3 + b 3 ) = (( ca 1 + b 1 )- ( ca 2 + b 2 ) , 2( ca 3 + b 3 )) = ( c ( a 1- a 2 ) + ( b 1- b 2 ) ,c (2 a 3 ) + 2 b 3 ) = c ( a 1- a 2 , 2 a 3 ) + ( b 1- b 2 , 2 b 3 ) = cT ( ) + T ( ) If { 1 , 2 , 3 } is a basis for R 3 , then the range is the span of { T ( 1 ) ,T ( 2 ) ,T ( 3 ) } . Take the standard ordered basis, { (1 , , 0) , (0 , 1 , 0) , (0 , , 1) } of R 3 . T (1 , , 0) = (1 , 0) T (0 , 1 , 0) = (- 1 , 0) T (0 , , 1) = (0 , 2) Hence, { (1 , 0) , (0 , 2) } is a basis for R T . In particular, dim R T = 2 = dim R 2 so T is onto. By the rank-nullity theorem, dim N T = 3- 2 = 1. It suffices to provide one non-zero vector in the kernel. We will use linearity of T . Note T (1 , , 0) + T (0 , 1 , 0) = (1 , 0) + (- 1 , 0) = (0 , 0). In particular, { (1 , 1 , 0) } is a basis for N T . Since there is a non-trivial kernel, T is not one-to-one. 1 (b) T : R 2 R 3 defined by T ( a 1 ,a 2 ) = ( a 1 + a 2 , , 2 a 1- a 2 ). Solution. Let = ( a 1 ,a 2 ) and = ( b 1 ,b 2 ) be two vectors in R 2 and c R any scalar. T ( c + ) = T ( ca 1 + b 1 ,ca 2 + b 2 ) = (( ca 1 + b 1 ) + ( ca 2 + b 2 ) , , 2( ca 1 + b 1 )- ( ca 2 + b 2 )) = ( c ( a 1 + a 2 ) + ( b 1 + b 2 ) , ,c (2 a 1- a 2 ) + (2 b 1- b 2 )) = c ( a 1 + a 2 , , 2 a 1- a 2 ) + ( b 1 + b 2 , , 2 b 1- b 2 ) = cT ( ) + T ( ) If { 1 , 2 } is a basis for R 2 , then the range is the span of { T ( 1 ) ,T ( 2 ) } . Take the standard ordered basis, { (1 , 0) , (0 , 1) } of R 2 . T (1 , 0) = (1 , , 2) T (0 , 1) = (1 , ,- 1) Hence, { (1 , , 2) , (1 , ,- 1) } is a basis for R T . In particular, dim R T = 2 6 = dim R 3 so T is not onto. By the rank-nullity theorem, dim N T = 2- 2 = 0 so that N T = { } and T is one-to-one. 2. Let T : V W be a linear transformation. Prove that ker T V is a subspace. Proof. Let , ker T , i.e., T ( ) = 0 = T ( ). For c F , we would like to show that c + ker T , i.e., T ( c + ) = 0. Since T : V W is linear: T ( c + ) = cT ( ) + T ( ) = c 0 + 0 = 0 3. Prove that a linear transformation T : V W is one-to-one if and only if ker T = { } ....
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235FA10-SAMPLEM2-sol - MTH 235 : Linear Algebra Sample...

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