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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 onetoone? 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 ( cα + 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 , 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 ranknullity theorem, dim N T = 3 2 = 1. It suffices to provide one nonzero 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 nontrivial kernel, T is not onetoone. 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 α = ( a 1 ,a 2 ) and β = ( b 1 ,b 2 ) be two vectors in R 2 and c ∈ R any scalar. T ( cα + β ) = 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 ( 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 , , 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 ranknullity theorem, dim N T = 2 2 = 0 so that N T = { } and T is onetoone. 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 cα + β ∈ ker T , i.e., T ( cα + β ) = 0. Since T : V → W is linear: T ( cα + β ) = cT ( α ) + T ( β ) = c · 0 + 0 = 0 3. Prove that a linear transformation T : V → W is onetoone if and only if ker T = { } ....
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This note was uploaded on 11/14/2010 for the course PHYCS 498 taught by Professor Aa during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
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