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University of Toronto Department of Mathematics MAT224H1S Linear Algebra I Midterm Exam I February 14, 2014 M. Mota, S. Uppal, F. Vera-Pacheco Duration: 1 hour 50 minutes Last Name: Given Name: Student Number: Email (@mail.utoronto.ca): Instructions : No calculators or other aids are allowed . Show all your work and justify your answers . If T : V W is a linear transformation, and α is a basis for V , and β is a basis for W , then [ T ] βα (also denoted by [ T ] β α in the textbook), is the matrix of T with respect to α and β You may use the back of each page for rough work but all your answers must be written on the front of each page . 1 of 14
[10] 1. Let α = { 1 , 1 + x, 1 + x 2 } , and β = { 1 , 1 + x, 1 + x + x 2 } be bases for P 2 ( C ). Let T : P 2 ( C ) P 2 ( C ) be the linear transformation such that [ T ] βα = i - i i 1 1 0 0 0 - 1 . Find T ( p ( x )) for every p ( x ) P 2 ( C ) .
and therefore [ T ( a + bx + cx 2 )] γ = 1 + i - 2 i - 2 1 0 - 2 0 0 - 1 a b c = (1 + i ) a - 2 ib - 2 c a - 2 c - c So, T ( p ( x )) = (1 + i ) a - 2 ib - 2 c + ( a - 2 c ) x - cx 2 . 3 of 14
2. Let T : R 3 M 2 × 2 ( R ) be the linear transformation defined by T ( a, b, c ) = a 5 a c 3 c . Consider the bases α = { (0 , 1 , 0) , (0 , 1 , 1) , (1 , 1 , 0) } of R 3 , and β = { 1 - 1 0 0 , 0 1 - 1 0 , 0 0 1 1 , 0 1 0 - 1 } of M 2 × 2 ( R ). [6] (a) Find [ T ] βα .
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by the above calculations, we have [ T ] γα = 0 0 1 0 0 5 0 1 0 0 3 0 . Using change of basis, [ T ] βα = [ I ] βγ [ T ] γα [ I ] αα = 1 2 2 0 0 0 1 1 - 1 1 1 1 1 1 1 1 1 - 1 0 0 1 0 0 5 0 1 0 0 3 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 = 0 0 1 0 1 3 0 2 3 0 - 1 3 .