hw1s - CME302 Homework 1 Solutions September 30, 2009 2. x...

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Unformatted text preview: CME302 Homework 1 Solutions September 30, 2009 2. x + y W = W (x + y) = W x + W y W x + W y = x the inequality following from the second axiom. 3. x W = W (x) = (W x) outside the norm using axiom 3. T&B 3.3a By definition, 1/2 W + y W, T&B 2.1 Suppose A Cmm is unitary and upper triangular (u.t.). The lower triangular (l.t.) case follows by symmetry of the problem. We show A is in fact diagonal: 1. By definition, A-1 = A . 2. The inverse of a nonsingular u.t. matrix U is also u.t. Consider solving U vi = ei , where ei is the ith column of the identity matrix. For a particular i, only the first i components of vi are nonzero (as evident in the back substitution algorithm). By construction, V = (v1 |v2 | |vm ) = U -1 . 3. The Hermitian transpose of a u.t. matrix is l.t.. 4. A matrix that is both u.t. (by 2) and l.t. (by 3) is diagonal. T&B 2.3a We show that the eigenvalues of a Hermitian matrix are real. First, (x Ax) = x A x = x Ax as A = A. If C is such that = , then the imaginary part is zero (written Im() = 0) and so in fact R. Consequently, both x Ax and x x are real scalars. If x is an eigenvector of A, then by definition Ax = x, and so x Ax = x x. As x Ax, x x R, so too is . T&B 2.3b We show that the eigenvectors of a Hermitian matrix are orthogonal. Let Ax = x and Ay = y; = by assumption. First, (Ay) = y A = y A = y as A is Hermitian. Next, premultiplying by y , y Ax = y x. The l.h.s. (short for "left hand side") is (y A)x = y x, and so ( - )y x = 0. As = by assumption, y x = 0: y is orthogonal to x. T&B 2.6 By the definition of the inverse of a matrix, AA-1 = I and so I = (I + uv )(I + uv ) = I + (1 + + v u)uv . The second term must be zero; therefore, 1++v u = 0, and so = (1+v u)-1 . A is singular if the denominator of is zero, which happens if v u = -1. In that case, Au = (I + uv )u = u + (v u)u = u - u = 0, and so u N (A). In fact, u is a basis for N (A), as the null space has dimension one. For suppose a vector w = 0 is not parallel to u. Then Aw = (I + uv )w = w + (v w)u = 0, and so w N (A). / T&B 3.1 To show that W W () is a vector norm, we must verify the three axioms of a vector norm, (3.1) in T&B: 1. As W is nonsingular, x = 0 iff (short for "if and only if") W x = 0. Therefore, x W = W x 0 with equality only if x = 0. 1 = || W x = || x W. was moved x Now, = max |xi | and i x 2 = i |xi | 2 . 1/2 i |xi | 2 max |xi |2 i 1/2 = max |xi |. i Equality is achieved if x = ej , for then max |xi | = || and i 1/2 i |xi |2 = ||2 = ||. T&B 3.3b We have 1/2 x 2 = i |xi | 2 m max |xi |2 i 1/2 = m max |xi | = m x . Equality is achieved if x is such that |xi | = R. For then 1/2 x x 2 = i i 2 = m and = max xi = , 2 and so x = m x . T&B 5.3a Let i be the eigenvalues of A A. Let the SVD of A be written U V . As A A = V T U U V = V T V = V diag()V , where diag() is a diagonal matrix having on its diagonal, i = i . For our problem, A= -2 11 -10 5 and A A = 104 -72 . -72 146 2 10 v1 10 v u 1 2 1 Solving det(A A - I) = (104 - )(146 - ) - 72 = 0 yields the eigenvalues = {200, 50}, and so = {10 2, 5 2}. Next, we find vi such that (A - i I)vi = 0 and vi 2 = 1. For 2 = 50, the first row of the system 104 - 50 -72 -72 146 - 50 a b u 2 2 implies 54a - 72b = 0, and so b = 54a/72. As the matrix has rank 1, the second row provides no additional information. Consequently, v2 2 v2 = 72 54 or, normalizing, v2 = 4/5 . 3/5 Figure 1: The unit circle (scaled by 10) and the image of the unit circle under the map A. T&B 5.3d We have A-1 = (U V )-1 = V - -1 U -1 = V -1 U = 1/20 -11/100 . 1/10 -1/50 The same procedure applied to 1 = 200 yields v2 . Assembling what we have found so far, -3/5 4/5 10 2 , A = U V = U 4/5 3/5 5 2 where we use a blank space to indicate 0, a useful convention. It remains to find U . We simply calculate 2 1 1 -1 U = AV = . 2 1 -1 T&B 5.3b The left singular vectors are ui ; the singular values, i ; and the right singular vectors, vi . See Figure 1. The unit circle is scaled by 10 for the illustration. T&B 5.3c Various norms of A are obtained as follows: A A A A 1 2 T&B 5.3e The eigenvalues i of A are determined by solving the quadratic equation implied by det(A - i I) = 0 as we did earlier for det(A A - i I) = 0: 3 i 391 . i = 2 T&B 5.3f Let = i 391. det A = (-2)(5) - (-10)(11) = 100 1 2 = (3 + )(3 - )/4 = 100 1 2 = (10 2)(5 2) = 100. = max A(:,j) j = 1 = 10 2 1/2 1 = 16 T&B 5.3g The area of an ellipse is ab, where a and b are half the semi-major and -minor axes. As V and U preserve the unit hypersphere, we need consider only the map by . Consequently, the unit circle is mapped to an ellipsoid having axes of length 2i , and so the area of the ellipse is 1 2 = 100. F = i i 2 i = 5 10 1 = max A(i,:) = 15. 2 ...
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This document was uploaded on 06/17/2010.

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