20105ee103_1_hw2_sols

# 20105ee103_1_hw2_sols - L Vandenberghe EE103 Homework 2...

This preview shows pages 1–3. Sign up to view the full content.

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 different order. For example, x 2 x 3 x 1 = 0 1 0 0 0 1 1 0 0 x 1 x 2 x 3 . (b) From the definition of permutation matrix, we see that A T is also a permutation matrix. Moreover A T A = I , because ( A T A ) ij = n summationdisplay k =1 a ki a kj = braceleftBigg 0 i negationslash = j 1 i = j. (If i negationslash = j , then a ki a kj = 0 for all k because there is exactly one entry equal to one in row k . If i negationslash = 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 A T A = a 2 2 + a 2 3 a 1 a 2 a 1 a 3 a 1 a 2 a 2 1 + a 2 3 a 2 a 3 a 1 a 3 a 2 a 3 a 2 1 + a 2 3 = a 2 1 + a 2 2 + a 2 3 0 0 0 a 2 1 + a 2 2 + a 2 3 0 0 0 a 2 1 + a 2 2 + a 2 3 a 2 1 a 1 a 2 a 1 a 3 a 1 a 2 a 2 2 a 2 a 3 a 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern