Linear Algebra Solutions 34

Linear Algebra Solutions 34 - 1 X 1 + (-1) X 2 + 0 X 3 + +...

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However if X 1 ,...,X m were linearly dependent, then one of these vec- tors would be a linear combination of the remaining vectors. Consequently S would be spanned by m - 1 vectors. But there exist a family of m lin- early independent vectors in S . Then by Theorem 3.3.2, we would have the contradiction m m - 1. 12. Let [ x, y, z ] t S . Then x + 2 y + 3 z = 0. Hence x = - 2 y - 3 z and x y z = - 2 y - 3 z y z = y - 2 1 0 + z - 3 0 1 . Hence [ - 2 , 1 , 0] t and [ - 3 , 0 , 1] t form a basis for S . Next ( - 1) + 2( - 1) + 3(1) = 0, so [ - 1 , - 1 , 1] t S . To Fnd a basis for S which includes [ - 1 , - 1 , 1] t , we note that [ - 2 , 1 , 0] t is not a multiple of [ - 1 , - 1 , 1] t . Hence we have found a linearly independent family of two vectors in S , a subspace of dimension equal to 2. Consequently these two vectors form a basis for S . 13. Without loss of generality, suppose that X 1 = X 2 . Then we have the non–trivial dependency relation:
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Unformatted text preview: 1 X 1 + (-1) X 2 + 0 X 3 + + 0 X m = 0 . 14. (a) Suppose that X m +1 is a linear combination of X 1 ,...,X m . Then h X 1 ,...,X m , X m +1 i = h X 1 ,...,X m i and hence dim h X 1 ,...,X m , X m +1 i = dim h X 1 ,...,X m i . (b) Suppose that X m +1 is not a linear combination of X 1 ,...,X m . If not all of X 1 ,...,X m are zero, there will be a subfamily X c 1 ,...,X c r which is a basis for h X 1 ,...,X m i . Then as X m +1 is not a linear combination of X c 1 ,...,X c r , it follows that X c 1 ,...,X c r , X m +1 are linearly independent. Also h X 1 ,...,X m , X m +1 i = h X c 1 ,...,X c r , X m +1 i . Consequently dim h X 1 ,...,X m , X m +1 i = r + 1 = dim h X 1 ,...,X m i + 1 . 37...
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