MATH311ansHW10

MATH311ansHW10 - Math 311, Homework 10 partial answers and...

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Unformatted text preview: Math 311, Homework 10 partial answers and solutions 3.7B.3. We apply the Gram-Schmidt procedure. Let v 1 = (1 , 2 , 1 , 1) . Since k (1 , 2 , 1 , 1) k 2 = 7 and h (1 , 2 , 1 , 1) , (- 1 , , 1 , 0) i = 0 , we can take v 2 = (- 1 , , 1 , 0) . Since k (- 1 , , 1 , 0) k 2 = 2 , h (1 , 2 , 1 , 1) , (0 , 1 , , 2) i = 4 , and h (- 1 , , 1 , 0) , (0 , 1 , , 2) i = 0 , we can take v 3 = (0 , 1 , , 2)- 4 7 (1 , 2 , 1 , 1) =- 4 7 ,- 1 7 ,- 4 7 , 10 7 . Since we are only asked to find an orthogonal and not an orthonormal basis, we can multiply this vector by 7 and take instead v 3 = (- 4 ,- 1 ,- 4 , 10) . For the fourth vector we start with any simple vector, for example (1 , , , 0) . Since k (- 4 ,- 1 ,- 4 , 10) k 2 = 133 , h (1 , 2 , 1 , 1) , (1 , , , 0) i = 1 , h (- 1 , , 1 , 0) , (1 , , , 0) i =- 1 , and h- 4 ,- 1 ,- 4 , 10) , (1 , , , 0) i =- 4 , we can take v 00 4 = (1 , , , 0)- 1 7 (1 , 2 , 1 , 1)-- 1 2 (- 1 , , 1 , 0)-- 4 133 (- 4 ,- 1 ,- 4 , 10) , or better, multiplying through by 266 , we can take instead...
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MATH311ansHW10 - Math 311, Homework 10 partial answers and...

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