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Linear Algebra Solutions 38

Linear Algebra Solutions 38 - a 11 x 1 ·· a r 1 x r = a...

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where A = [ a ij ] is non–singular. Then the above system of equations can be solved for X 1 , . . . , X r in terms of Y 1 , . . . , Y r . Consequently by Theorem 3.2.1 h Y 1 , . . . , Y r i = h X 1 , . . . , X r i = S. It follows from problem 11 that Y 1 , . . . , Y r is a basis for S . (ii) We show that all bases for S are given by equations 2. So suppose that Y 1 , . . . , Y r forms a basis for S . Then because X 1 , . . . , X r form a basis for S , we can express Y 1 , . . . , Y r in terms of X 1 , . . . , X r as in 2, for some matrix A = [ a ij ]. We show A is non–singular by demonstrating that the linear independence of Y 1 , . . . , Y r implies that the rows of A are linearly independent. So assume x 1 [ a 11 , . . . , a 1 r ] + · · · + x r [ a r 1 , . . . , a rr ] = [0 , . . . ,
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Unformatted text preview: a 11 x 1 + ··· + a r 1 x r = . . . a 1 r x 1 + ··· + a rr x r = . Hence x 1 Y 1 + ··· + x r Y r = x 1 ( a 11 X 1 + ··· + a 1 r X r ) + ··· + x r ( a r 1 X 1 + ··· + a rr X r ) = ( a 11 x 1 + ··· + a r 1 x r ) X 1 + ··· + ( a 1 r x 1 + ··· + a rr x r ) X r = X 1 + ··· + 0 X r = 0 . Then the linear independence of Y 1 ,...,Y r implies x 1 = 0 ,...,x r = 0. (We mention that the last argument is reversible and provides an alter-native proof of part (i).) 41...
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