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# hw9solutions - Homework 9 Solutions Joshua Hernandez...

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Homework 9 Solutions Joshua Hernandez December 6, 2009 6.3 - The Adjoint of a Linear Operator 2b. Let V = C 2 (over C ) and linear transformation g : V F deﬁned by the mapping g ( z 1 ,z 2 ) = z 1 - 2 z 2 , ﬁnd a vector y such that g ( x ) = h x,y i for all x V : Solution: Clearly, y = (1 , -2) works. Taking the standard orthonormal basis { e i } n i =1 , y = n X i =1 g ( e i ) e i = e 1 - 2 e 2 = (1 , -2) . 3b. Let V = C 2 (over C ) and linear operators T on V , let T : V V be deﬁned by the mapping T ( z 1 ,z 2 ) = (2 z 1 + iz 2 , (1 - i ) z 1 ) . Evaluate T * at x = (3 - i, 1 + 2 i ). Solution: Let α be the standard orthonormal basis of C 2 . [ T * x ] α = [ T ] * α [ x ] α = ± 2 i 1 - i 0 ² * ± 3 - i 1 + 2 i ² = ± 2 1 + i - i 0 ²± 3 - i 1 + 2 i ² = ± 5 + i 1 - 3 i ² . 6. Let T be a linear operator on an inner product space V . let U 1 = T + T * and U 2 = TT * . Prove that U 1 = U * 1 and U 2 = U * 2 . Solution: In two lines, U * 1 = ( T + T * ) * = T * + ( T * ) * = T * + T = U 1 , U * 2 = ( TT * ) * = ( T * ) * ( T ) * = TT * = U 2 . 12. Let V be an inner product space, and let T be a linear operator on V . Prove the following results. a. R ( T * ) = N ( T ) 1

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Solution: z R ( T * ) ⇐⇒ h T * ( x ) ,z i = h x, T ( z ) i = 0 for all x V ⇐⇒ z N ( T ) . The right equivalence is a consequence of inner product property (d). b. If V is ﬁnite-dimensional, then R ( T * ) = N ( T ) . Solution: Above, we proved that R ( T ) = N ( T ). Taking the perpendicular of both sides, ( R ( T * ) ) = N ( T ) . By problem (6.2.13c) in the last homework, ( R ( T * ) ) = R ( T * ). Note that, to apply that result, we only needed rank( T * ) rank( T ) < . 18. Let A be an n × n matrix. Prove that det( A * ) = det( A ). Solution: This is trivial if A = ( a ) M 1 × 1 : det( A * ) = det( ( ¯ a ) ) = ¯ a = det( ( a ) ) = det( A ) . Assume the result is true for ( n - 1) × ( n - 1) matrices. Then, det( f A * ij ) = det(( e A ji ) * ) = (det( e A ji )) . Expanding along the ﬁrst column, det( A * ) = ( A * ) 11 det( f A * 11 ) + ··· + (-1) n +1 ( A * ) n 1 det( f A * n 1 ) = A 11 det( e A 11 ) + ··· + (-1) n +1 A 1 n det( e A 1 n ) , which is just the conjugate of det( A ), expanded along the ﬁrst row. 20c. Use the least squares approximation to ﬁnd the best ﬁts with both (i) a linear function and (ii) a quadratic function. Compute the error E in both cases.
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## This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.

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hw9solutions - Homework 9 Solutions Joshua Hernandez...

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