20105ee103_1_hw2_sols

20105ee103_1_hw2_sols - L. Vandenberghe EE103 10/7/10...

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L. Vandenberghe 10/7/10 EE103 Homework 2 solutions 1. Exercise 2.1. A = I 10 × 10 B 0 10 × 10 B T 0 5 × 5 0 5 × 10 0 10 × 10 0 10 × 5 BB T . A is 25 × 25. 2. Exercise 2.4. (a) y = f ( x ) is a permutation of the elements of x , i.e. , the elements of y are the elements of x in a diFerent order. ±or example, x 2 x 3 x 1 = 0 1 0 0 0 1 1 0 0 x 1 x 2 x 3 . (b) ±rom the de²nition of permutation matrix, we see that A T is also a permutation matrix. Moreover A T A = I , because ( A T A ) ij = n s k =1 a ki a kj = b 0 i n = j 1 i = j. (If i n = j , then a ki a kj = 0 for all k because there is exactly one entry equal to one in row k . If i n = j , then a ki a kj = 0 for all k except the k for which a ki = 1.) Therefore g ( f ( x )) = A T Ax = x for all x . In other words, g ( x ) = A T x is the inverse permutation of the permuta- tion f ( x ) = Ax . It is the permutation that maps y = f ( x ) back to x . In the example, y 3 y 1 y 2 = 0 0 1 1 0 0 0 1 0 y 1 y 2 y 3 . If we apply this permutation to ( y 1 , y 2 , y 3 ) = ( x 2 , x 3 , x 1 ), we obtain ( x 1 , x 2 , x 3 ) again. 1
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3. Exercise 2.5. (a) a × x = Ax with A = 0 a 3 a 2 a 3 0 a 1 a 2 a 1 0 . (b) Working out the matrix product gives
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20105ee103_1_hw2_sols - L. Vandenberghe EE103 10/7/10...

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