204PS5Solution2009

204PS5Solution2009 - Econ 204 Problem Set 5 Solutions...

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Econ 204 Problem Set 5 Solutions Exercise 1 C is already diagonal. B can be diagonalized: it has eigenvalues 1 and 2 and a pair of eigenvectors (1 ; 0) and (1 ; 1) corresponding to these eigenvalues. Let V = f (1 ; 0) , (1 ; 1) g be the set with the two eigenvectors. It is also a basis for R 2 . Let W be the standard basis and consider the change of basis matrix Mtx V;W ( id ) = Mtx W;V ( id ) ° 1 = ° 1 1 0 1 ± ° 1 . Thus, B can be written as B = Mtx W ( B ) = Mtx W;V ( id ) ° Mtx V ( B ) ° Mtx V;W ( id ) where Mtx V ( B ) is precisely the diagonal matrix ° 1 0 0 2 ± (since (1 ; 0) and (1 ; 1) are both the eigenvectors of B and the basis vectors in V ). To check that this works, Mtx W;V ( id ) ° Mtx V ( B ) ° Mtx V;W ( id ) = ° 1 1 0 1 ± ° 1 0 0 2 ± ° 1 1 0 1 ± ° 1 = ° 1 1 0 1 ± ° 1 0 0 2 ± ° 1 ± 1 0 1 ± = ° 1 2 0 2 ± ° 1 ± 1 0 1 ± = ° 1 1 0 2 ± = B: Eigenvalues for matrix A are 1 ; ± 1 ; and 2 with eigenvectors (1 ; 0 ; ± 1) , (0 ; 0 ; 1) , and (2 ; 1 ; ± 2) . Let M the matrix formed by these eigenvectors and K the di- agonal matrix with eigenvalues on the diagonal. You can proceed similarly for matrix A and verify that A = MKM ° 1 ; M ° 1 is thus again the change of basis matrix from the standard basis to the basis formed by the above eigenvectors. Exercise 2 If A is positive semide°nite and B is an n by m matrix, then B T AB is indeed positive semide°nite. Let x be a vector in R m . Then Bx is a vector in R n and since A is positive semide°nite, x T B T ABx = ( Bx ) T A ( Bx ) ² 0 ; so that B T AB is positive semide°nite. Now suppose A is positive de°nite. Previous reasoning shows that x T B T ABx ² 0 ; however we need the inequality to be strict for x 6 = 0 . Since A is positive de°nite, x T B T ABx 6 = 0 for x 6 = 0 if and only if Bx 6
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