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tut04s - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS Linear Mathematics 2012 Tutorial 4 (week 5) — Solutions 1. In each of the following find a basis for Span( X ) , and say whether or not Span( X ) is equal to the vector space R n of which X is a subset. a ) X = ( 1 5 ) , ( 2 6 ) , ( 3 7 ) , ( 4 8 ) R 2 . b ) X = 1 2 3 , 4 5 6 , 0 0 0 R 3 . c ) X = 1 2 3 4 , 2 4 6 8 , 0 0 0 0 R 4 . d ) X = 1 1 1 1 1 , - 1 - 1 - 1 - 1 - 1 , 4 4 4 4 4 R 5 . Solution a ) Since { ( 1 5 ) , ( 2 6 ) } is linearly independent, and dim( R 2 ) = 2 , { ( 1 5 ) , ( 2 6 ) } is a basis for Span( X ) , and Span( X ) = R 2 . (Indeed, any subset of X containing two vectors is linearly independent, and is a basis for Span( X ) , as is the standard basis { ( 1 0 ) , ( 0 1 ) } .) b ) Span( X ) = a 1 2 3 + b 4 5 6 + c 0 0 0 | a, b, c R = a 1 2 3 + b 4 5 6 | a, b R , and 1 2 3 , 4 5 6 is linearly independent. Hence this set is a basis for Span( X ) , which therefore has dimension 2, and is not equal to R 3 . c ) The vectors 2 4 6 8 and 0 0 0 0 are scalar multiples of 1 2 3 4 , and so a basis for Span( X ) is 1 2 3 4 . Span( X ) therefore has dimension 1, and is not equal to R 4 . d ) The vectors - 1 - 1 - 1 - 1 - 1 and 4 4 4 4 4 are scalar multiples of 1 1 1 1 , and so a basis for Span( X ) is 1 1 1 1 1 . Span( X ) therefore has dimension 1, and is not equal to R 5 . 2. Find a basis for the subspace V = x y z R 3 2 x + y = 3 z of R 3 . Solution An arbitrary element x y z of V can be written in the form x y z = x 3 z - 2 x z , since x y z V if and only if y = 3 z - 2 x . Now, x 3 z - 2 x z = x 1 - 2 0 + z 0 3 1 , showing that 1 - 2 0 , 0 3 1 spans V . Since 1 - 2 0 is not a scalar multiple of 0 3 1 , this set is clearly linearly independent and is a basis for V , which is a subspace of R 3 of dimension 2 . Math 2061: Tutorial 4 (week 5) — Solutions S.B and A.M 20/3/2012
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Linear Mathematics Tutorial 4 (week 5) — Solutions Page 2 3. Let X = 6 3 2 , 2 3 1 , 4 - 6 0 . Find the dimension of Span( X ) . Solution Note that the subset containing the first two vectors of X is linearly independent (since the vectors are not scalar multiples of each other), and so the dimension of Span( X ) is at least 2. Hence, the dimension is either 2 or 3, depending on whether or not X is linearly independent. Suppose that a 6 3 2 + b 2 3 1 + c 4 - 6 0 = 0 0 0 , for some a, b, c R . To find all possible a, b, c we have to solve the system of linear equations corresponding to the following augmented matrix: 6 2 4 0 3 3 - 6 0 2 1 0 0 Row reduce -----→ 1 0 2 0 0 1 - 4 0 0 0 0 0 . (Note that the row echelon form of a matrix is not unique, as it depends upon the sequence of row operations that you choose; the reduced row echelon form is, however, unique. Consequently, you might obtain a different row echelon form for the augmented matrix above. However, your row echelon form should still give the same set of solutions as the one above. This comment applies to all of the questions below which compute row echelon forms.)
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