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hw4_solutions

# hw4_solutions - EE263s Summer 2009-10 Laurent Lessard...

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EE263s Summer 2009-10 Laurent Lessard EE263s homework 4 1. Orthogonal matrices. (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. 1

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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. 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. Projection matrices. A matrix P R n × n is called a projection matrix if P = P T and P 2 = P . (a) Show that if P is a projection matrix then so is I P . (b) Suppose that the columns of U R n × k are orthonormal. Show that UU T is a projection matrix. (Later we will show that the converse is true: every projection matrix can be expressed as UU T for some U with orthonormal columns.) (c) Suppose A R n × k is full rank, with k n . Show that A ( A T A ) - 1 A T is a projection matrix. (d) If S R n and x R n , the point y in S closest to x is called the projection of x on S . Show that if P is a projection matrix, then y = Px is the projection of x on R ( P ). (Which is why such matrices are called projection matrices . . . ) Solution.
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