Linear Algebra Solutions 33

Linear Algebra Solutions 33 - , 1 , 1], (that is at least...

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form which we deduce that x 1 = 0 ,...,x m = 0. Secondly we have to prove that every vector of S is expressible as a linear combination of X 1 , X 1 + X 2 ,...,X 1 + ··· + X m . Suppose X S . Then X = a 1 X 1 + ··· + a m X m . We have to Fnd x 1 ,...,x m such that X = x 1 X 1 + x 2 ( X 1 + X 2 ) + ··· + x m ( X 1 + ··· + X m ) = ( x 1 + x 2 + ··· + x m ) X 1 + ··· + x m X m . Then a 1 X 1 + ··· + a m X m = ( x 1 + x 2 + ··· + x m ) X 1 + ··· + x m X m . So if we can solve the system x 1 + x 2 + ··· + x m = a 1 ,...,x m = a m , we are Fnished. Clearly these equations have the unique solution x 1 = a 1 - a 2 ,...,x m - 1 = a m - a m - 1 , x m = a m . 10. Let A = a b c 1 1 1 . If [ a, b, c ] is a multiple of [1 , 1 , 1], (that is, a = b = c ), then rank A = 1. ±or if [ a, b, c ] = t [1 , 1 , 1] , then R ( A ) = h [ a, b, c ] , [1 , 1 , 1] i = h t [1 , 1 , 1] , [1 , 1 , 1] i = h [1 , 1 , 1] i , so [1 , 1 , 1] is a basis for R ( A ). However if [ a, b, c ] is not a multiple of [1
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Unformatted text preview: , 1 , 1], (that is at least two of a, b, c are distinct), then the lefttoright test shows that [ a, b, c ] and [1 , 1 , 1] are linearly independent and hence form a basis for R ( A ). Conse-quently rank A = 2 in this case. 11. Let S be a subspace of F n with dim S = m . Also suppose that X 1 ,...,X m are vectors in S such that S = h X 1 ,...,X m i . We have to prove that X 1 ,...,X m form a basis for S ; in other words, we must prove that X 1 ,...,X m are linearly independent. 36...
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