s02 - Ma 450: Mathematics for Multimedia Solution: to...

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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 N-cube, the unit cube in Euclidean N-space. How may other vertices are connected to it by single edges? Solution: Imagine aligning the N-cube with the coordinate axes in Euclidean N-space 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 m-dimensional subspace of an N-dimensional 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 finite-dimensional, 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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s02 - Ma 450: Mathematics for Multimedia Solution: to...

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