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# solhw2 - Solutions for Homework 2 Numerical Linear Algebra...

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Solutions for Homework 2 Numerical Linear Algebra 1 Fall 2011 Problem 2.1 Let x C n and y C n be two arbitrary vectors. Consider determining a circulant matrix C C n × n such that y = Cx 2.1.a . Assume that C exists for a given pair ( x, y ), show how to construct it. 2.1.b . When is C unique for a given pair ( x, y )? 2.1.c . When does C not exist for a given pair ( x, y )? Solution: There are two equivalent lines of reasoning. The first is in the “spatial” domain. Denote C = C ( c ) to indicate that it is generated by its first row c T . We have Cx = y C ( c ) x = y H ( x ) c = y C ( x r ) c r = y where H ( x ) C n × n . H ( x ) is not however circulant in the same fashion as C ( c ). It is a Hankel matrix not a Toeplitz matrix. A Hankel matrix is constant along its antidiagonals. It is therefore circulant with respect to the first row using a circulant shift to the left (not to the right as with C ( c )) and with respect to the first column using a circulant shift up (not down as with C ( c )). The Hankel/left circulant form of the matrix vector product can be transformed into the Toeplitz/right circulant form of the matrix vector product. This is done by using a Toeplitz/right circulant matrix and a righthand side vector x r and c r that are the vectors x and c with their elements in reverse order respectively, i.e., P T = ( e n e n - 1 . . . e 1 ) , x r = Px, c r = Pc Since x and y are given, one way to find c is to solve the system with the matrix, H ( x ), defined by x . Therefore, to have a solution we must have y ∈R ( H ( x )) or y ∈R ( C ( x r )). 1

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These facts are easily seen by example with n = 4. x = ξ 0 ξ 1 ξ 2 ξ 3 c = γ 0 γ 1 γ 2 γ 3 y = C ( c ) x = γ 0 ξ 0 + γ 1 ξ 1 + γ 2 ξ 2 + γ 3 ξ 3 γ 3 ξ 0 + γ 0 ξ 1 + γ 1 ξ 2 + γ 2 ξ 3 γ 2 ξ 0 + γ 3 ξ 1 + γ 0 ξ 2 + γ 1 ξ 3 γ 1 ξ 0 + γ 2 ξ 1 + γ 3 ξ 2 + γ 0 ξ 3 = γ 0 ξ 0 + γ 1 ξ 1 + γ 2 ξ 2 + γ 3 ξ 3 γ 0 ξ 1 + γ 1 ξ 2 + γ 2 ξ 3 + γ 3 ξ 0 γ 0 ξ 2 + γ 1 ξ 3 + γ 2 ξ 0 + γ 3 ξ 1 γ 0 ξ 3 + γ 1 ξ 0 + γ 2 ξ 1 + γ 3 ξ 2 = H ( § ) c = γ 3 ξ 3 + γ 2 ξ 2 + γ 1 ξ 1 + γ 0 ξ 0 γ 3 ξ 0 + γ 2 ξ 3 + γ 1 ξ 2 + γ 0 ξ 1 γ 3 ξ 1 + γ 2 ξ 0 + γ 1 ξ 3 + γ 0 ξ 2 γ 3 ξ 2 + γ 2 ξ 1 + γ 1 ξ 0 + γ 0 ξ 3 = C ( x r ) c r The system of course can be solved via the DFT which gives the more elegant view of the problem in the Fourier domain (essentially the eigendomain).
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solhw2 - Solutions for Homework 2 Numerical Linear Algebra...

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