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Linear Algebra Solutions 35

Linear Algebra Solutions 35 - N A = nullity A = n-rank A =...

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Our result can be rephrased in a form suitable for the second part of the problem: dim h X 1 , . . . , X m , X m +1 i = dim h X 1 , . . . , X m i if and only if X m +1 is a linear combination of X 1 , . . . , X m . If X = [ x 1 , . . . , x n ] t , then AX = B is equivalent to B = x 1 A * 1 + · · · + x n A * n . So AX = B is soluble for X if and only if B is a linear combination of the columns of A , that is B C ( A ). However by the first part of this question, B C ( A ) if and only if dim C ([ A | B ]) = dim C ( A ), that is, rank [ A | B ] = rank A . 15. Let a 1 , . . . , a n be elements of F , not all zero. Let S denote the set of vectors [ x 1 , . . . , x n ] t , where x 1 , . . . , x n satisfy a 1 x 1 + · · · + a n x n = 0 . Then S = N ( A ), where A is the row matrix [ a 1 , . . . , a n ]. Now rank A = 1 as A 6 = 0. So by the “rank + nullity” theorem, noting that the number of columns of A equals n , we have
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Unformatted text preview: N ( A ) = nullity( A ) = n-rank A = n-1 . 16. (a) (Proof of Lemma 3.2.1) Suppose that each of X 1 ,...,X r is a linear combination of Y 1 ,...,Y s . Then X i = s X j =1 a ij Y j , (1 ≤ i ≤ r ) . Now let X = ∑ r i =1 x i X i be a linear combination of X 1 ,...,X r . Then X = x 1 ( a 11 Y 1 + ··· + a 1 s Y s ) + ··· + x r ( a r 1 Y 1 + ··· + a rs Y s ) = y 1 Y 1 + ··· + y s Y s , where y j = a 1 j x 1 + ··· + a rj x r . Hence X is a linear combination of Y 1 ,...,Y s . Another way of stating Lemma 3.2.1 is h X 1 ,...,X r i ⊆ h Y 1 ,...,Y s i , (1) 38...
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