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APSC 174J, Solutions #8 Posted: May 19, 2016 Section 11, Problem 1(b). The linear transformation determined by A is LA : R → R̂2 , defined by −x LA (x) = 2x −1 −1 (1) The Range Im(A) is the linear span of { }. Since is a non-zero vector, 2 2 it is linearly independent, hence a basis for Im(A). Therefore, dim(Im(A))=1. (2) To find the Kernel of A, we solve for LA (x) = 0: −x 0 = 2x 0 the solution is x = 0. Hence Ker(A)={0}, and dim(Ker(A))=0. (3) Rank(A)=dim(Im(A))=1, Nullity(A)=dim(Ker(A))=0. The rank-nullity theorem is satisfied: 1 + 0 = 1 = the number of columns of A Section 11, Problem 1(d). The linear transformation determined by A is LA : R̂3 → R̂2 , defined by x1 3x + 2x2 + x3 ) = 1 LA ( x 2 0 x3 3 2 1 (1) The Range Im(A) is the linear span of { , , }. Easy to see that 0 0 0 3 1 2 1 1 = 3 and = 2 , hence a basis for Im(A) is { }. Therefore, 0 0 0 0 0 dim(Im(A))=1. 2 (2) To find the Kernel of A, we solve for LA (x) = 0: 0 3x + 2x2 + x3 = 1 0 0 the solution is x3 = −3x1 − 2x2 and x1 , x2 can be any real numbers. Write the solution in parametric vector form we have x1 x1 1 0 = x1 0 + x2 1 x= x2 x2 = x3 −3x1 − 2x2 −3 −2 1 0 1 0 , 1 }. Check that { 0 , 1 } is So Ker(A) is the linear span of { 0 −3 −2 −3 −2 0 1 linearly independent. Hence ( 0 , 1 ) is a basis for Ker(A), and −2 −3 dim(Ker(A))=2. (3) Rank(A)=dim(Im(A))=1, Nullity(A)=dim(Ker(A))=2. The rank-nullity theorem is satisfied: 1 + 2 = 3 = the number of columns of A Section 11, Problem 1(e). The linear transformation determined by A is LA : R̂2 → R̂2 , defined by x1 0 LA ( ) = x2 0 so LA is the zero transformation. 0 0 (1) The Range Im(A) is the linear span of { , }, i.e. Im(A)= {0} Therefore, 0 0 dim(Im(A))=0. (2) To find the Kernel of A, we solve for LA (x) = 0. Since LA (x) = 0 for any x ∈ R̂2 , we have Ker(A)= R̂2 , and dim(Ker(A))=2. 3 (3) Rank(A)=dim(Im(A))=0, Nullity(A)=dim(Ker(A))=2. The rank-nullity theorem is satisfied: 0 + 2 = 2 = the number of columns of A Section 11, Problem 1(r). The linear transformation determined by A is LA : R̂4 → R̂2 , defined by x1 x 3x2 2 LA ( ) = x3 −x 3 x4 0 3 0 0 (1) The Range Im(A) is the linear span of { , , , , }. Easy to see 0 0 −1 0 3 0 that a basis for Im(A) is { , , }. Therefore, dim(Im(A))=2. 0 −1 (2) To find the Kernel of A, we solve for LA (x) = 0: 0 3x 2 = 0 −x3 the solution is x2 = x3 = 0 and x1 , x4 can be any real numbers. Write the solution in parametric vector form we have x1 x1 1 0 x 0 0 0 2 x = = = x1 + x4 x3 0 0 0 x4 x4 0 1 1 0 1 0 0 0 0 0 So Ker(A) is the linear span of { , }. Check that { , } is linearly 0 0 0 0 0 1 0 1 4 1 0 0 0 independent. Hence ( , ) is a basis for Ker(A), and dim(Ker(A))=2. 0 0 0 1 (3) Rank(A)=dim(Im(A))=1, Nullity(A)=dim(Ker(A))=2. The rank-nullity theorem is satisfied: 2 + 2 = 4 = the number of columns of A Section 11, Problem 1(z). The linear transformation determined by A is LA : R̂4 → R̂3 , defined by x1 x + x + 2x − x 1 2 3 4 x 2 LA ( ) = 3x + 6x + 3x 1 3 4 x3 −2x1 − 4x3 + 2x2 x4 −1 2 1 (1) The Range Im(A) is the linear span of { 3 , 0 , 6 , 3 }. To find a 2 −4 0 −2 basis of Im(A), we reduce A to row-echelon form: 1 −1 1 1 2 −1 1 1 2 −1 3 0 6 → 0 −3 0 6 → 0 −3 0 6 3 −2 0 −4 2 0 2 0 0 0 0 0 4 1 1 2 Columns with leading arethe first, second and fourth column. Thus a entries 1 1 −1 basis for Im(A) is { 3 , 0 , 3 }. Therefore, dim(Im(A))=3. 2 0 −2 (2) To find the Kernel of A, we solve for LA (x) = 0: x1 + x2 + 2x3 − x4 0 3x1 + 6x3 + 3x4 = 0 −2x1 − 4x3 + 2x2 0 5 1 1 2 3 0 6 −2 0 −4 −1 0 3 2 1 1 2 −1 0 0 −3 0 → 0 0 0 0 0 6 4 0 0 the solution is x4 = 0, x2 = 2x4 = 0, x1 = −2x3 and x3 is free. Write the solution in parametric vector form we have x1 −2x3 −2 x 0 0 2 x= = = x3 x3 x3 1 x4 0 0 −2 −2 0 0 So Ker(A) is the linear span of { }. Since is a non-zero vector, it is 1 1 0 0 −2 0 linearly independent. Hence ( ) is a basis for Ker(A), and dim(Ker(A))=1. 1 0 (3) Rank(A)=dim(Im(A))=3, Nullity(A)=dim(Ker(A))=1. The rank-nullity theorem is satisfied: 3 + 1 = 4 = the number of columns of A