Mathematics_1_oneside.pdf

85 prove theorem 88 draw a picture that illustrates

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8.5 Prove Theorem 8.8 . Draw a picture that illustrates property (iii). H INT : Use Theorems 8.2 and 8.7 . 8.6 Let x R n be a non-zero vector. Show that x k x k is a normalized vector. Is the condition x 6= 0 necessary? Why? Why not? 8.7 (a) Show that k x k 1 and k x k satisfy the properties of a norm. (b) Draw the unit balls in R 2 , i.e., the sets { x R 2 : k x k ≤ 1 } with respect to the norms k x k 1 , k x k 2 , and k x k . (c) Use a computer algebra system of your choice (e.g., Maxima ) and draw unit balls with respect to the p -norm for various values of p . What do you observe? 8.8 Prove Theorem 8.14 . Draw a picture that illustrates property (iii). H INT : Use Theorem 8.8 . 8.9 Show that k x + y k 2 -k x - y k 2 = 4 x 0 y . H INT : Use k x k 2 = x 0 x . 8.10 Prove Theorem 8.18 . H INT : Use k x k 2 = x 0 x . 8.11 Prove Theorem 8.22 . H INT : Represent x by means of c ( x ) and compute x 0 v j . 8.12 Let U = a b c d . Give conditions for the elements a , b , c , d that im- ply that U is an orthogonal matrix. Give an example for such an orthogonal matrix.
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9 Projections To them, I said, the truth would be literally nothing but the shadows of the images. Suppose we are given a subspace U R n and a vector y R n . We want to find a vector u U such that the “error” r = y - u is as small as possible. This procedure is of great importance when we want to reduce the num- ber of dimensions in our model without loosing too much information. 9.1 Orthogonal Projection We first look at the simplest case U = span( x ) where x R n is some fixed normalized vector, i.e., k x k = 1. Then every u U can be written as λ x for some λ R . Let y , x R n be fixed with k x k = 1. Let r R n and λ R such that Lemma 9.1 y = λ x + r . Then for λ = λ * and r = r * the following statements are equivalent: (1) k r * k is minimal among all values for λ and r . (2) x 0 r * = 0. (3) λ * = x 0 y . P ROOF . (2) (3): Follows by a simple computation (see Problem 9.1 ). An immediate consequence is that there always exist r * and λ * such that r * = y - λ * x and x 0 r * = 0. (2) (1): Assume that x 0 r * = 0 and λ * such that r * = y - λ * x . Set r ( ε ) = y - ( λ * + ε ) x = ( y - λ * x ) - ε x = r * - ε x for ε R . As r * and x are orthogonal by our assumption the Pythagorean theorem implies k r ( ε ) k 2 = k r * k 2 + ε 2 k x k 2 = k r * k 2 + ε 2 . Thus k r ( ε ) k ≥ k r * k where equality holds if and only if ε = 0. Thus r * minimizes k r k . 63
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9.1 O RTHOGONAL P ROJECTION 64 (1) (2): Assume that r * minimizes k r k and λ * such that r * = y - λ * x . Set r ( ε ) = y - ( λ * + ε ) x = r * - ε x for ε R . Our assumption implies that k r * k 2 ≤ k r * - ε x k 2 = k r * k 2 - 2 ε x 0 r + ε 2 k x k 2 for all ε R . Thus 2 ε x 0 r ε 2 k x k 2 = ε 2 . Since ε may have positive and negative sign we find - ε 2 x 0 r ε 2 for all ε 0 and hence x 0 r = 0, as claimed. Orthogonal projection. Let x , y R n be two vectors with k x k = 1. Then Definition 9.2 p x ( y ) = ( x 0 y ) x is called the orthogonal projection of y onto the linear span of x .
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