Linear Algebra Solutions 36

Linear Algebra Solutions 36 - ,...,Y s are linearly...

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if each of X 1 ,...,X r is a linear combination of Y 1 ,...,Y s . (b) (Proof of Theorem 3.2.1) Suppose that each of X 1 ,...,X r is a linear combination of Y 1 ,...,Y s and that each of Y 1 ,...,Y s is a linear combination of X 1 ,...,X r . Then by (a) equation (1) above h X 1 ,...,X r i ⊆ h Y 1 ,...,Y s i and h Y 1 ,...,Y s i ⊆ h X 1 ,...,X r i . Hence h X 1 ,...,X r i = h Y 1 ,...,Y s i . (c) (Proof of Corollary 3.2.1) Suppose that each of Z 1 ,...,Z t is a linear combination of X 1 ,...,X r . Then each of X 1 ,...,X r , Z 1 ,...,Z t is a linear combination of X 1 ,...,X r . Also each of X 1 ,...,X r is a linear combination of X 1 ,...,X r , Z 1 ,...,Z t , so by Theorem 3.2.1 h X 1 ,...,X r , Z 1 ,...,Z t i = h X 1 ,...,X r i . (d) (Proof of Theorem 3.3.2) Let Y 1 ,...,Y s be vectors in h X 1 ,...,X r i and assume that s > r . We have to prove that Y 1
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Unformatted text preview: ,...,Y s are linearly dependent. So we consider the equation x 1 Y 1 + + x s Y s = 0 . Now Y i = r j =1 a ij X j , for 1 i s . Hence x 1 Y 1 + + x s Y s = x 1 ( a 11 X 1 + + a 1 r X r ) + + x r ( a s 1 X 1 + + a sr X r ) . = y 1 X 1 + + y r X r , (1) where y j = a 1 j x 1 + + a sj x s . However the homogeneous system y 1 = 0 , , y r = 0 has a nontrivial solution x 1 ,...,x s , as s > r and from (1), this results in a nontrivial solution of the equation x 1 Y 1 + + x s Y s = 0 . 39...
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