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Midterm

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Department
Mathematics
Course
MAT224H1
Professor
Sean Uppal
Semester
Fall

Description
University of Toronto Department of Mathematics MAT224H1F Linear Algebra II Midterm Examination October 13, 2010 S. Uppal Duration: 1 hour 20 minutes Last Name: Given Name: Student Number: Tutorial Code: No calculators or other aids are allowed. FOR MARKER USE ONLY Question Mark 1 /10 2 /10 3 /10 4 /10 5 /5 6 /5 TOTAL /50 1 of 10 1. Find a basis for the kernel and the range of the linear transformation T: C ! C 3 whose matrix with respect to the standard basis of C is 0 1 2 i 1 + i @ A 1 ▯ i 1 2 : 0 1 ▯ i 2 Solution: We ▯nd the row reduced echelon form of the given matrix. 0 1 0 1 0 1 2 i 1 + i 2 !2 2▯1 2 i 1 + i r3$2 2 i 1 + i @ 1 ▯ i 1 2 A ▯▯▯▯▯▯▯! @0 0 0 A ▯▯▯! @ 0 1 ▯ i 2 A 0 1 ▯ i 2 0 1 ▯ i 2 0 0 0 1 0 1 i 2 0 2 1 0 1 1 !1 1▯i @ A @ A ▯▯▯▯▯▯▯▯▯! 0 1 ▯ i 2 ▯▯▯▯! 0 1 i + 1 0 0 0 0 0 0 Hence the kernel is spanned by f(1;i + 1;▯1)g and a basis for the range is the set the ▯rst two columns of the matrix for T, that is the basis 0 1 0 1 ( 2 i ) @ 1 ▯ i; 1 A 0 1 ▯ i 2 of 10 EXTRA PAGE FOR QUESTION 1 - do not remove. 3 of 10 2. Let T: P2(R) ! P 2R) be de▯ned by T(p(x)) = p(ax+b), where a;b 2 R are constants, a 6= 0;1. Find a basis ▯ for P (R) such that [T]is diagonal. 2 ▯▯ Solution: We begin by ▯nding the eigenvalues and eigenvectors for the matrix of T with respect to the standard basis ▯. We have that T(1) = 1, T(x) = ax+b and T(x ) = (ax+b) = 2 2 2 a x + 2abx + b hence 0 1 1 b b2 @ A [T]▯▯ = 0 a 2ab 0 0 a 2 2 We have that det([T▯▯ ▯▯I) = (1▯▯)(a▯▯)(a ▯▯) so the eigenvalues are ▯ 1 1, ▯ 2 a and ▯ = a : To get the eigenvectors we solve the system of equations corresponding 3 to the following matrices (corresponding to1▯ 2▯ 3▯ respectively) 0 2 1 0 2 1 0 2 2 1 0 b b 1 ▯ a b b 1 ▯ a b b @0 a ▯ 1 2ab A ; 0 0 2ab A ; 0 a ▯ a2 2abA 2 2 0 0 a ▯ 1 0 0 a ▯ a 0 0 0 b ▯b2 2b we get the eigenvectors 1 = (0;0;1), 2 = (a▯1;1;0) and ((1▯a); 1▯a;▯1). If we take P to be the matrix with vector i as column i then we have that ▯1 2 P [T]▯▯ = diagf1;a;a g: Hence we can take ▯ to be the basis 2 b ▯b2 2b 2 ▯ = fx ; + x; 2 + x ▯ x g a ▯ 1 (1 ▯ a) 1 ▯ a 4 of 10 EXTRA PAGE FOR QUESTION 2 - do not remove. 5 of 10 3 2 3. Let V = f(x ;x1;x2) 23R j x +x +x1= 0g2and 3 = fa+bx+cx 2 P (R) j a▯b = 2 2cg. Show that V and W are isomorphic and construct an isomorphism T: V ! W. Justify your answer. Solution: Suppose that x = (x ;x ;x1) 2 V 3 Then x = (x 1x ;2 )3 = (x 1x ;2(x + 1 )) 2 = x 11;0;▯1) + x (0;2;▯1) showing that a basis for V is f(1;0;▯1);(0;1;▯1)g and so dim(V ) = 2 Similarly if p(x) = a + bx + cx 2 W then 2 p(x) = a + bx + cx a ▯ b 2 = a + bx + x 2 1 2 1 2 = a(1 + x ) + b(x ▯ x ) 2 2 showing that a basis for W is f1 + x2;x ▯ x2g and so dim(W) = 2. Since V and W 2 2 have the same dimension and are ▯nite dimensional they are isomorphic. An explicit 1 2 1 2 isomorphism is given by T(x ;x ;1 ) 2 x3(1 + 1 2x ) + x 2x ▯ x 2. We need to show that this is a linear function which is one -to -one and onto. To see the function is linear note that if ▯ 2 R and x;y 2 V then 1
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