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Unformatted text preview: Solutions to Homework #4, MAT1341A Question # 1: (2 × 6 = 12 points) For each of the following, decide if the given set B is a basis of the corresponding vector space V . In each case, justify your answer. You may refer to any theorems discussed in class. (a) V = R 3 , B = { (1 , 2 , 3) , (4 , 5 , 6) , (7 , 8 , 9) , (1 , , 1) } . Solution: We know that dim( R 3 ) = 3, but B has 4 vectors in it. Therefore, by the theorem in class, B cannot possibly be linearly independent. Thus B is not a basis of V . (b) V = M 2 , 2 ( R ), B = 1 2 3 4 , 1 2 3 , 1 4 5 . Solution: We know that dim( M 2 , 2 ( R )) = 4, but B has 3 vectors in it. Therefore, by the theorem in class, B cannot possibly span all of M 2 , 2 ( R ). Thus B is not a basis of V . (c) V = R 3 , B = { (1 , 2 , 3) , (4 , 5 , 6) , (2 , 4 , 6) } . Solution: In this case, B has the right number of vectors, that is, 3 = dim( V ). But the first and third vectors are scalar multiples of one another, so this set is clearly linearly dependent, and so is not a basis. (d) V = P 3 , B = { 1 + x, x + x 2 , x 2 x 3 } Solution: We know that dim( P 3 ) = 4, but this set has only 3 vectors in it; so as above, it cannot possibly span all of V and thus is not a basis. (e) V = P 1 , B = { 1 + x, 2 3 x } Solution: Since dim( P 1 ) = 2 and B has two elements, it suffices by the theorem shown in class to verify that B is linearly independent. Since there are just two vectors, and they are not scalar multiples of one another, B is LI. Since B consists of 2 LI elements in a 2dimensional space, it’s a basis....
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This note was uploaded on 09/28/2011 for the course MATH 1341 taught by Professor Daigle during the Winter '10 term at University of Ottawa.
 Winter '10
 DAIGLE
 Vector Space

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