hw5_sol_sej

hw5_sol_sej - EE103 Spring 2010, HW 5 Solution Set, Prof...

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EE103 Spring 2010, HW 5 Solution Set, Prof S.E. Jacobsen Page 1 / 5 Applied Numerical Computing Instructor: Prof. S. E. Jacobsen HW5 Solution Set Problem 1. (a) Since P is orthogonal, we have T PP I , and we have PA LU . Then, we have TT T A ULP (1) Then T Ay d can be expressed as y d (2) Or, we may rewrite (2) as () UL P We then have   , , T T T T zLP y Uz d p r o v i d e sz uP y Lu z prov idesu yP u (b) yrow s u . Students, you should generate a small example to see that this result is correct. Problem 2. (a) Since N is orthonormal, N T N = I . Then 22 T T T Nx Nx Nx x N Nx x Ix x x x  , (b) One may observe the L.H.S. of the equation, stated in the question , is 12 11 2 2 , , ..., , , ..., ( ... ) T kk k k T I qq q q Iq q q q q q     Since the orthogonality between q i and q j , 1 i , j k , gives q iT q j = 0, the R.H.S., of the equation, stated in the question , is
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EE103 Spring 2010, HW 5 Solution Set, Prof S.E. Jacobsen Page 2 / 5 (1 )(1 ) 22 11 ) 22 11 ) ( 1 )( 1 ) 1 ( )( )...( )( ) ( )( )...( ) ( )( )...( ) ... ( )( ... ) ( kk T k k T T T T k T T T T T k T T T T T k Iq q Iq q q Iq q I qq I q q qq qq qq q q q q   12 2 ( 1 ) ( 1 ) 2 2 ) ( 1 ) ... ) ( ... ) TT k k T k k T k k T k k T q q q q q q q q q q  (c) Note that q k+1 can be decomposed as follows. ) ) 2 ) 1 2 1 ... kTk kT k kT k T T aq qa q q q If 2 0 q , then we see that 1 1 () T aa q q . Since 1 /( ) q a norm a , we would have that 2 a is a multiple of 1 a and thereby contradicting the linear independence of the columns of A . Continue in this fashion. That is, q k+1 = 0 would imply that a k+1 is a linear combination of ( q 1 , q 2 , …, q k
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This note was uploaded on 06/01/2010 for the course EE EE 103 taught by Professor Jacobsen during the Spring '09 term at UCLA.

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hw5_sol_sej - EE103 Spring 2010, HW 5 Solution Set, Prof...

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