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235FA10-SAMPLEM2-sol

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

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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 α = ( a 1 , a 2 , a 3 ) and β = ( b 1 , b 2 , b 3 ) be two vectors in R 3 and c R any scalar. T ( + β ) = 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 ( α ) + T ( β ) If { α 1 , α 2 , α 3 } is a basis for R 3 , then the range is the span of { T ( α 1 ) , T ( α 2 ) , T ( α 3 ) } . Take the standard ordered basis, { (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1) } of R 3 . T (1 , 0 , 0) = (1 , 0) T (0 , 1 , 0) = ( - 1 , 0) T (0 , 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 , 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

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(b) T : R 2 R 3 defined by T ( a 1 , a 2 ) = ( a 1 + a 2 , 0 , 2 a 1 - a 2 ). Solution. Let α = ( a 1 , a 2 ) and β = ( b 1 , b 2 ) be two vectors in R 2 and c R any scalar. T ( + β ) = T ( ca 1 + b 1 , ca 2 + b 2 ) = (( ca 1 + b 1 ) + ( ca 2 + b 2 ) , 0 , 2( ca 1 + b 1 ) - ( ca 2 + b 2 )) = ( c ( a 1 + a 2 ) + ( b 1 + b 2 ) , 0 , c (2 a 1 - a 2 ) + (2 b 1 - b 2 )) = c ( a 1 + a 2 , 0 , 2 a 1 - a 2 ) + ( b 1 + b 2 , 0 , 2 b 1 - b 2 ) = cT ( α ) + T ( β ) If { α 1 , α 2 } is a basis for R 2 , then the range is the span of { T ( α 1 ) , T ( α 2 ) } . Take the standard ordered basis, { (1 , 0) , (0 , 1) } of R 2 . T (1 , 0) = (1 , 0 , 2) T (0 , 1) = (1 , 0 , - 1) Hence, { (1 , 0 , 2) , (1 , 0 , - 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 = { 0 } 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 α, β ker T , i.e., T ( α ) = 0 = T ( β ). For c F , we would like to show that + β ker T , i.e., T ( + β ) = 0. Since T : V W is linear: T ( + β ) = cT ( α ) + T ( β ) = c · 0 + 0 = 0 3. Prove that a linear transformation T : V W is one-to-one if and only if ker T = { 0 } . Proof. ” Suppose that the linear transformation T : V W is one-to-one. Suppose that α ker T , then by definition, T ( α ) = 0. In particular, T (0) = 0 = T ( α ). Since T assumed to be one-to-one, α = 0. ” Suppose ker T = { 0 } . If T ( α ) = T ( β ), then T ( α ) - T ( β ) = 0. Since T linear, T ( α - β ) = 0. Since the kernel is trivial, α - β = 0 so that α = β .
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