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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 2010 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 Therefore, one way to solve the problem is to solve the system with the matrix deﬁned by x . H ( x ) is not however circulant in the same fashion as C ( c ). It is a Hankel matrix not Toeplitz. 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. It by using a Toeplitz/right circulant matrix with the vectors must be reversed, i.e., x r and c r are the vectors x and c with their elements in reverse order respectively. Therefore, to have a solution we must have y ∈ R ( H ( x )) or y ∈ R ( C ( x r )). 1

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This is 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 ( x ) 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
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solhw2 - Solutions for Homework 2 Numerical Linear Algebra...

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