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20095ee103_1_hw2_sols

# 20095ee103_1_hw2_sols - L Vandenberghe EE103 Homework 2...

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L. Vandenberghe 10/8/09 EE103 Homework 2 solutions 1. Exercise 2.3. (a) Ax is a circular downward shift over one position: Ax = 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 x 1 x 2 x 3 x 4 x 5 = x 5 x 1 x 2 x 3 x 4 . (b) A k is a circular shift over k positions. A 2 = 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 , A 3 = 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 , A 4 = 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 , and A 5 = I . (If we shift x over five positions, we obtain x again.) (c) A T x shifts x in the other direction: A T x = 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 x 1 x 2 x 3 x 4 x 5 = x 2 x 3 x 3 x 4 x 1 . A T is also the inverse of A , because A T A = I . 2. Exercise 2.5. f is linear. x T y is a scalar, so ( x T y ) y = y ( x T y ) = y ( y T x ) = ( yy T ) x. In the first step we write the multiplication of y with the scalar x T y as a matrix-matrix multiplication of a matrix y (with one column) and a 1 × 1-matrix x T y . In the second step we use the fact that x T y = y T x . In the last step, we use the fact that matrix multiplication is associative.

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20095ee103_1_hw2_sols - L Vandenberghe EE103 Homework 2...

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