MidTermSoln

# MidTermSoln - MATH 136 Winter 2008 Midterm Examination...

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MATH 136 Winter 2008 Midterm Examination: Solutions February 13, 2008 [10] Question 1: Mark each statement True or False. No justiﬁcation necessary. 1 mark for each correct answer, - 1 2 for each wrong answer and 0 for no answer. (i) Let A be a 4 × 3 real matrix. Then Col A is a subspace of R 4 . True ± False ± (ii) If A is any 5 × 3 matrix and ~ b is any vector in R 5 such that the equation A~x = ~ b has a solution, then the solution is unique. True ± False ± (iii) Let V be a vector space over a ﬁeld F . If a~x = b~x for a, b F and ~x V , then a = b . True ± False ± (iv) If f and g are polynomials of degree 100 over rationals, then f + g is a polynomial of degree 100. True ± False ± (v) The span of three vectors in a vector space V is a subspace of V . True ± False ± (vi) The set of real numbers is a vector space over the ﬁeld of rational numbers. True ± False ± (vii) Let V = C 2 = { ( x, y ) : x, y C } and W = { ( x, y ) V : x 2 + y 2 = 0 } . Then W is a subspace of V . True ± False ± ~v 1 = (1 , i ) , ~v 2 = ( - 1 , i ) gives ~v 1 + ~v 2 = (0 , 2 i ) / W . (viii) Let T : R 2 R 2 be given by T ( u 1 , u 2 ) = ± ( u 1 , u 2 ) if u 1 6 = 0 (0 , 0) if u 1 = 0 . Then T is a linear transformation. True ± False ± ~v 1 = (0 , 1) , ~v 2 = (1 , 1) gives T ( ~v 1 + ~v 2 ) 6 = T ( ~v 1 ) + T ( ~v 2 ). (ix) Let T : R 3 R be given by T ( u 1 , u 2 , u 3 ) = u 1 + u 2 + u 3 . Then T is a linear transformation. True ± False ± (x) Let T : R n R be such that T ( ~u ) =largest component of ~u . Then T is a linear transformation. True ± False ± ~v 1 = (1 , 2 , ··· , n ) ,~v 2 = ( - 1 , - 2 , ··· , - n ) gives T ( ~v 1 ) = n, T ( ~v 2 ) = - 1 but T ( ~v 1 + ~v 2 ) = 0.

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MATH 136 Winter 2008 Midterm Examination: Solutions Page 2 [6] Question 2: You are given M 2 × 2 ( R ), the set of all 2 × 2 matrices over R ; P 3 ( R ), the set of all polynomials of degree 3 over R together with the zero polynomial, and R 4 . Standard rules for matrix, polynomial and vector addition apply, as does the standard rule for scalar multiplication. Scalars are drawn from R . [2]( i ) Is A = ± 3 4 0 2 ² in Span ³ B = ± 1 1 0 1 ² , C = ± 1 0 0 1 ² , D = ± 0 0 0 1 ²´ ? If yes, express A as a linear combi- nation of B, C and D . If not, explain why not. Yes . Let b, c, d be such that b ± 1 1 0 1 ² + c ± 1 0 0 1 ² + d ± 0 0 0 1 ² = ± 3 4 0 2 ² or ± b + c b 0 b + c + d ² = ± 3 4 0 2 ² . Then b = 4 , b + c = 3 , b + c + d = 2 giving b = 4 , c = d = - 1. Hence A = 4 B - C - D . ± [2](
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## This note was uploaded on 01/13/2011 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.

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MidTermSoln - MATH 136 Winter 2008 Midterm Examination...

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