{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw3_2011_04_16_01_solutions

# hw3_2011_04_16_01_solutions - EE263 S Lall 2011.04.16.01...

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

EE263 S. Lall 2011.04.16.01 Homework 3 Solutions Due Thursday 4/21. 1. Orthogonal matrices. 44015 (a) Show that if U and V are orthogonal, then so is UV . (b) Show that if U is orthogonal, then so is U - 1 . (c) Suppose that U R 2 × 2 is orthogonal. Show that U is either a rotation or a reflection. Make clear how you decide whether a given orthogonal U is a rotation or reflection. Solution. (a) To prove that UV is orthogonal we have to show that ( UV ) T ( UV ) = I given U T U = I and V T V = I . We have ( UV ) T ( UV ) = V T U T UV = V T V (since U T U = I ) = I (since V T V = I ) and we are done. (b) Since U is square and orthogonal we have U - 1 = U T and therefore by taking inverses of both sides U = ( U T ) - 1 or equivalently U = ( U - 1 ) T (the inverse and transpose operations commute.) But U T U = I and by substitution U - 1 ( U - 1 ) T = I . Since U - 1 is square this also implies that ( U - 1 ) T U - 1 = I so U - 1 is orthogonal. (c) Suppose that U = bracketleftbigg a b c d bracketrightbigg R 2 × 2 is orthogonal. This is true if and only if columns of U are of unit length, i.e. , a 2 + c 2 = 1 and b 2 + d 2 = 1, columns of U are orthogonal, i.e. , ab + cd = 0. Since a 2 + c 2 = 1 we can take a and c as the cosine and sine of an angle α respectively, i.e. , a = cos α and c = sin α . For a similar reason, we can take b = sin β and d = cos β . Now ab + cd = 0 becomes cos α sin β + sin α cos β = 0 or sin( α + β ) = 0 . The sine of an angle is zero if and only if the angle is an integer multiple of π . So α + β = or β = α with k Z . Therefore U = bracketleftbigg cos α sin( α ) sin α cos( α ) bracketrightbigg . Now two things can happen: k is even so sin( α ) = sin α and cos( α ) = cos α , and therefore U = bracketleftbigg cos α sin α sin α cos α bracketrightbigg . Clearly, from the lecture notes, this represents a rotation. Note that in this case det U = cos 2 α + sin 2 α = 1. k is odd so sin( α ) = sin α and cos( α ) = cos α , and therefore U = bracketleftbigg cos α sin α sin α cos α bracketrightbigg . From the lecture notes, this represents a reflection. The determinant in this case is det U = cos 2 α sin 2 α = 1. 1

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

View Full Document
EE263 S. Lall 2011.04.16.01 Therefore we have shown that any orthogonal matrix in R 2 × 2 is either a rotation or reflection whether its determinant is +1 or 1 respectively. 2. Some properties of the product of two matrices. 43025 For each of the following statements, either show that it is true, or give a (specific) counterexample. If AB is full rank then A and B are full rank. If A and B are full rank then AB is full rank. If A and B have zero nullspace, then so does AB . If A and B are onto, then so is AB . You can assume that A R m × n and B R n × p . Some of the false statements above become true under certain assumptions on the dimensions of A and B . As a trivial example, all of the statements above are true when A and B are scalars, i.e. , n = m = p = 1. For each of the statements above, find conditions on n , m , and p that make them true. Try to find the most general conditions you can. You can give your conditions as inequalities involving n , m , and p
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