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Unformatted text preview: + c n d n , and k ~x k 2 = c 2 1 + · · · + c 2 n . Solution: We have < ~x,~ y > = < c 1 ~v 1 + · · · + c n ~v n , d 1 ~v 1 + · · · + d n ~v n > = c 1 < ~v 1 , d 1 ~v 1 + · · · + d n ~v n > + · · · + c n < ~v n , d 1 ~v 1 + · · · + d n ~v n > = c 1 d 1 < ~v 1 ,~v 1 > + · · · + c 1 d n < ~v 1 ,~v n > + c 2 d 1 < ~v 2 ,~v 1 > + c 2 d 2 < ~v 2 ,~v 2 > + · · · + · · · + c 2 d n < ~v 2 ,~v n > + · · · + c n d 1 < ~v n ,~v 1 > + · · · + c n d n < ~v n ,~v n > But, since { ~v 1 , . . . ,~v n } is orthonormal we have < ~v i ,~v i > = 1 and < ~v i ,~v j > = 0 for i 6 = j , hence we get < ~x,~ y > = c 1 d 1 + · · · + c n d n , as required. Now observe that by taking d i = c i for all i we get y = x and obtain k ~x k 2 = < ~x,~x > = c 2 1 + · · · + c 2 n , as required....
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.
 Spring '10
 WILKIE
 Matrices

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