Linear Algebra Solutions 32

Linear Algebra Solutions 32 - S . We have to prove that X 1...

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In Section 1.6, problem 17, we found that A had reduced row–echelon form B = 1 0 0 0 0 1 0 b 0 0 1 1 . From B we read o± the following: (a) The rows of B form a basis for R ( A ). (Consequently the rows of A also form a basis for R ( A ). (b) The ²rst three columns of A form a basis for C ( A ). (c) To ²nd a basis for N ( A ), we solve AX = 0 and equivalently BX = 0. From B we see that the solution is x 1 = 0 x 2 = - bx 4 = bx 4 x 3 = - x 4 = x 4 , where x 4 is an arbitrary element of F . Hence X = x 4 0 b 1 1 , so [0 , b, 1 , 1] t is a basis for N ( A ). 9. Suppose that X 1 ,...,X m form a basis for a subspace
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Unformatted text preview: S . We have to prove that X 1 , X 1 + X 2 ,...,X 1 + + X m also form a basis for S . First we prove the independence of the family: Suppose x 1 X 1 + x 2 ( X 1 + X 2 ) + + x m ( X 1 + + X m ) = 0 . Then ( x 1 + x 2 + + x m ) X 1 + + x m X m = 0 . Then the linear independence of X 1 ,...,X m gives x 1 + x 2 + + x m = 0 ,...,x m = 0 , 35...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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