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hw1 Soln Rev SEJ

# hw1 Soln Rev SEJ - EE103 Spring 2010 HW1 Sol Prof S.E...

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EE103 Spring 2010 HW1 Sol. Prof. S.E. Jacobsen 1 Applied Numerical Computing EE103 Spring 2010 Instructor: Prof. S. E. Jacobsen HW1 Solution Problem 1 (i) Define = B A I . Since there exists a non-zero vector x such that B x = 0, one can deduce that the columns of B are linearly dependent. Therefore, by definition, ( - ) A I is singular. (ii) For n = 2, 11 12 21 22 11 12 21 22 11 22 12 21 2 2 11 22 11 22 12 21 det( ) det( ) det( ) ( )( ) ( 1) ( ) ( ) A I a a I a a a a a a a a a a a a a a a a   For n = 3, 11 12 13 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 22 23 21 23 21 22 11 12 13 32 33 31 33 31 32 det( ) det( ) det( ) ( )det( ) det( ) det( ) ( A I a a a a a a I a a a a a a a a a a a a a a a a a a a a a a a a a a a a 11 22 33 23 32 12 21 33 23 31 13 21 32 31 22 2 11 22 33 22 33 23 32 12 21 12 21 33 12 23 31 13 31 13 31 22 13 21 32 3 2 11 22 33 1 )[( )( ) ] [ ( ) ] [ ( )] ( )[ ( ) ( )] [ ] [ ] ( 1) ( ) ( a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a   1 22 11 33 22 33 23 32 12 21 13 31 11 22 33 11 23 32 12 21 33 12 23 31 13 31 22 13 21 32 ) ( ) a a a a a a a a a a a a a a a a a a a a a a a a a a a a a One can prove the claim by induction. Based on the case where n = 2, one can show the claim is true. Assume the claim holds for n = k . For n = k+ 1, define B A λ I and expand det ( A λ I ): 1 1 1 1 1 det( ) ( 1) k j j j j A I b M In the above expression we denote (a) b ij the entry at row i and column j of matrix B and (b) M ij the determinant of a k by k matrix that results from B by removing row i

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EE103 Spring 2010 HW1 Sol. Prof. S.E. Jacobsen 2 and column j . By induction, M 11 is a k -th order polynomial in λ with leading coefficient (-1) k . The rest of the determinants M 1j represent at most ( k-1 )-th order polynomials in λ . Notice that b 11 = a 11 λ . Therefore, det ( A λ I ) is a ( k +1)-th order polynomial in λ with leading coefficient (-1) k+1 .
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