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Unformatted text preview: Ma 450: Mathematics for Multimedia Solution: to Homework Assignment 2 Prof. Wickerhauser Due Friday, February 19th, 2010 1. Fix a vertex in the Ncube, the unit cube in Euclidean Nspace. How may other vertices are connected to it by single edges? Solution: Imagine aligning the Ncube with the coordinate axes in Euclidean Nspace in such a way that the chosen vertex is at the origin and the edges at that vertex lie in the positive rays of the N coordinate axes. The vertices connected to the origin by single edges will then lie at coordinate 1 in each of the N directions, so there will be exactly N of them. 2 2. Let P , Q , S be subspaces of R N with respective dimensions p, q, s . Suppose that S = P + Q . (a) Prove that max { p, q } ≤ s ≤ p + q . (b) Find an example that achieves the equality s = max { p, q } . (c) Find an example that achieves the equality s = p + q . Solution: (a) Let P = span { u 1 , . . . , u p } and Q = span { v 1 , . . . , v q } define bases. Then S = P + Q = span { u 1 , . . . , u p ; v 1 , . . . , v q } , so the dimension of S is at most p + q . Since P is a subspace of S , the dimension of S is at least p , the dimension of P . Similarly, s ≥ q . Thus s ≥ max { p, q } . (b) Use R 3 with P = span { e 1 } and Q = span { e 1 , e 2 } . (c) Use R 2 with P = span { e 1 } and Q = span { e 2 } . 2 3. Prove Inequality 2.15 for every N . Solution: Fix N and let x ∈ C N be any vector. Then  x ( k )  ≤ max { x ( i )  : i = 1 , . . . , N } = k x k ∞ for each k = 1 , . . . , N , so k x k 1 =  x (1)  + ··· +  x ( N )  ≤ N k x k ∞ . Also,  x ( k )  ≥ 0 for all k = 1 , . . . , N , so k x k 1 =  x (1)  + ··· +  x ( N )  ≥ max { x ( i )  : i = 1 , . . . , N } = k x k ∞ , proving the other inequality. Since x was arbitrary, the inequalites hold for all of R N . Since N was arbitrary, both equalities hold for any N . 2 4. Prove that k x y k ≥ k x k  k y k for any vectors x , y in a normed vector space X . Solution: Use the norm sublinearity axiom twice: k x k = k ( x y ) + y k ≤ k x y k + k y k ⇒ k x y k ≥ k x k  k y k ; k y k = k ( y x ) + x k ≤ k y x k + k x k ⇒ k y x k ≥ k y k  k x k . 1 But k y x k = k ( 1)( x y ) k =  1 k x y k = k x y k , so the two inequalities may be combined: k x y k ≥ max {k y k  k x k , k x k  k y k} = k x k  k y k , since max { z, z } =  z  for any real number z . 2 5. Suppose that Y is an mdimensional subspace of an Ndimensional inner product space X . Prove that Y ⊥ is at most N m dimensional. Solution: First check the trivial case: If m = 0, then Y = { } , so Y ⊥ = X is N = N 0 = N m dimensional. Otherwise, let V = { v 1 , . . . , v m } be a basis for Y = span V . Since Y ⊥ is a subspace of a finite dimensional space, it too is finitedimensional, so let W = { w 1 , . . . , w k } be its basis. The dimension of Y ⊥ is k , so it remains to show that m + k ≤ N ....
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This note was uploaded on 05/16/2010 for the course MATH 450 taught by Professor Wickerhauser during the Spring '10 term at Washington University in St. Louis.
 Spring '10
 Wickerhauser
 Math

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