orthog-proj

orthog-proj - Math 1b Prac Orthogonal bases, orthogonal...

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Math 1b Prac — Orthogonal bases, orthogonal projection January 29, 2009 — expanded February 11, 2009 [This is an outline of material I want to cover before starting Chapter 4 of the text. Examples and motivation will be given in class.] We will write h x , y i for the dot product of vectors x , y R n in this section. This is one type of inner product ; see page 307. Properties of inner products include h c a , b i = c h a , b i and h a 1 + a 2 , b i = h a 1 , b i + h a 2 , b i . Recall that two vectors x , y are orthogonal when h x , y i =0 . Wh enw esaytha t v 1 , v 2 ,..., v k are orthogonal, we mean that every pair v i and v j , i 6 = j , are orthogonal. The length of v is the square-root of the inner product of v with itself: || v || = p h v , v i . This is a positive number (and in particular, not zero) as long as v 6 = 0 .The (Euclidean) distance between u and w is the length of u w , i.e. || u w || . Note that if a and b are orthogonal, || a + b || 2 = || a || 2 + || b || 2 . We may call this the Phythagorean Theorem. It holds because || a + b || 2 = h a + b , a + b i = h a , a i + h a , b i + h b , a i + h b , b i = || a || 2 +0+0+ || b || 2 . Theorem 1. Nonzero orthogonal vectors v 1 , v 2 v k are linearly independent. Proof: Suppose v 1 , v 2 v k are nonzero orthogonal vectors and that c 1 v 1 + c 2 v 2 + ... + c k v k = 0 . Given i , take the inner product of both sides with v i to get c 1 h v 1 , v i i + + c i h v i , v i i + + c k h v k , v i i = h 0 , v i i . Most of the inner products are zero, and the only surviving term on the right is c i h v i , v i i = c i || v i || 2 , and because || v i || 2 6 = 0, it must be that c i = 0. This holds for all i =1 , 2 ,...,k ,andth is means That v 1 , v 2 v k are linearly independent. ut
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Theorem 2. For If u 1 , u 2 ,..., u k are nonzero orthogonal vectors and v is any vector, then v 0 = v ± h v , u 1 i h u 1 , u 1 i u 1 + ...
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orthog-proj - Math 1b Prac Orthogonal bases, orthogonal...

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