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quiz4-solns - = 0 But each of these terms is nonnegative so...

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MATH 110 - SOLUTION TO QUIZ 4 LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA ˜ NEZ Let V be an inner product space and suppose that ( e 1 , . . . , e m ) is an orthonormal list of vectors in V . For v V , prove that v span( e 1 , . . . , e m ) if and only if k v k 2 = |h v, e 1 i| 2 + · · · + |h v, e m i| 2 . Proof. First, extend ( e 1 , . . . , e m ) to an orthonormal basis ( e 1 , . . . , e m , e m +1 , . . . , e n ) of V . For any v V , we can write v = h v, e 1 i e 1 + · · · + h v, e n i e n . Using this, by the Pythagorean Theorem, we have k v k 2 = |h v, e 1 i| 2 + · · · + |h v, e m i| 2 + |h v, e m +1 i| 2 + · · · + |h v, e n i| 2 . Now, from this we see that k v k 2 = |h v, e 1 i| 2 + · · · + |h v, e m i| 2 if and only if |h v, e m +1 i| 2 + · · ·
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Unformatted text preview: = 0 . But each of these terms is nonnegative, so this is possible if and only if h v,e m +1 i = ··· = h v,e n i = 0 . Finally, using the expression for v given at the beginning, this is possible if and only if v = h v,e 1 i e 1 + ··· + h v,e m i e m . Thus we have shown that k v k 2 = |h v,e 1 i| 2 + ··· + |h v,e m i| 2 if and only if v is a linear combination of ( e 1 ,...,e m ), as was to be shown. ± Date : July 23, 2009....
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