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 ±nd a basis for Span( X ) ,andsaywhetherornot Span( X ) is equal to the vector space R n of which X is a subset. a ) X = n ( 1 5 ) , ( 2 6 ) , ( 3 7 ) , ( 4 8 ) o R 2 . b ) X = 1 2 3 ² , ± 4 5 6 ² , ± 0 0 0 ²o R 3 . c ) X = n ³ 1 2 3 4 ´ , ³ 2 4 6 8 ´ , ³ 0 0 0 0 ´ o R 4 . d ) X = 1 1 1 1 1 , µ - 1 - 1 - 1 - 1 - 1 , µ 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 ) ,asisthestandardbasis { ( 1 0 ) , ( 0 1 ) } .) b )Sp an ( X n a ± 1 2 3 ² + b ± 4 5 6 ² + c ± 0 0 0 ² | a, b, c R o = n a ± 1 2 3 ² + b ± 4 5 6 ² | a, b R o , and 1 2 3 ² , ± 4 5 6 ²o is linearly independent. Hence this set is a basis for Span( X ) ,wh ich therefore has dimension 2, and is not equal to R 3 . c ) The vectors ³ 2 4 6 8 ´ and ³ 0 0 0 0 ´ are scalar multiples of ³ 1 2 3 4 ´ ,andsoabas isfor Span( X ) is n ³ 1 2 3 4 ´ o . Span( X ) therefore has dimension 1, and is not equal to R 4 . d ) The vectors µ - 1 - 1 - 1 - 1 - 1 and µ 4 4 4 4 4 are scalar multiples of ³ 1 1 1 1 ´ ,andsoabasisfor 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 ² R 3 · · · 2 x + y =3 z o of R 3 . Solution An arbitrary element ± x y z ² of V can be written in the form ± x y z ² = ± x 3 z - 2 x z ² ,s ince ± x y z ² V if and only if y z - 2 x .N ow , ± x 3 z - 2 x z ² = x ± 1 - 2 0 ² + z ± 0 3 1 ² ,show ingtha t 1 - 2 0 ² , ± 0 3 1 ²o spans V .S ± 1 - 2 0 ² is not a scalar multiple of ± 0 3 1 ² ,thissetisclearlylinearly independent and is a basis for V ,whichisasubspaceof 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 , ± 2 3 1 ² , ± 4 - 6 0 ²o .F indthed imens ionof Span( X ) . Solution Note that the subset containing the ±rst 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 ² + b ± 2 3 1 ² + c ± 4 - 6 0 ² = ± 0 0 0 ² , for some a, b, c R . To ±nd all possible a, b, c we have to solve the system of linear equations corresponding to the following augmented matrix: ³ 62 4 0 33 - 6 0 21 0 0 ´ Row reduce -----→ ³ 10 2 0 01 - 4 0 00 0 0 ´ .
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tut04s - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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