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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- Fall '10
- Dreibelbis
- Linear Algebra, Linear Independence, Elementary algebra, XR, Yr
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