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Unformatted text preview: Math 235 Assignment 5 Due 9:15am, Wednesday Feb 28, 2007. 1. From the Text Â§ 5.3 #31. Construct a nonzero 2 Ã— 2 matrix that is invertible but not diag onalizable. Solution. Consider a matrix 1 1 1 . The determinant of this matrix is one, implying this matrix is invertible. It is easy to see that the matrix has only one eigenvalue Î» = 1 with algebraic multiplicity two. The eigenspace corresponding to the eigenvalue is spanned by 1 , thus geometric multiplicity of Î» is one which is less than algebraic multiplicity of Î» . Therefore the matrix is not diagonalizable. #32. Construct a nondiagonal 2 Ã— 2 matrix that is diagonalizable but not invertible. Solution. Consider 1 1 1 1 . The determinant of the matrix is zero, implying this matrix is not invertible. The characteristic equation of the matrix is given by Î» ( Î» 2) = 0 . Thus there are two distinct eigenvalues, Î» 1 = 0 and Î» 2 = 2 . Since the dimension of the matrix is also two, then the matrix is diagonalizable. Â§ 5.4 #10. Define T : P 3â†’ R 4 by T ( p ) = p ( 3) p ( 1) p (1) p (3) . a. Show that T is a linear transformation. b. Find the matrix for T relative to the basis { 1 , t, t 2 , t 3 } for P 3 and the standard basis for R 4 . Solution. a. We need to show that T ( c p + d q ) = cT ( p ) + dT ( q ) , where c and d are constants. 1 T ( c p + d q ) = ( c p + d q )( 3) ( c p + d q )( 1) ( c p + d q )(1) ( c p + d q )(3) = c ( p ( 3)) + d ( q ( 3)) c ( p ( 1)) + d ( q ( 1)) c ( p (1)) + d ( q (1)) c ( p (3)) + d ( q (3)) = c p ( 3) p ( 1) p (1) p (3) + d q ( 3) q ( 1) q (1) q (3) = cT ( p ) + dT ( q ) b. Let B = { 1 , t, t 2 , t 3 . } Then p ( t ) = a + bt + ct 2 + dt 3 = a b c d B . We want to find a matrix A such that A a b c d B = p ( 3) p ( 1) p (1) p (3) = a + b ( 3) + c ( 3) 2 + d ( 3) 3 a + b ( 1) + c ( 1) 2 + d ( 1) 3 a + b (1) + c (1) 2 + d (1) 3 a + b (3) + c (3) 2 + d (3) 3 . It is easy to see that A = 1 3 9 27 1 1 1 1 1 1 1 1 1 3 9 27 . Â§ 5.5 #4. Let the following matrix act on C 2 . Find the eigenvalues and a basis for each eigenspace in C 2 . 5 2 1 3 . Solution. 5 Î» 2 1 3 Î» = (5 Î» )(3 Î» ) + 2 = Î» 2 8 Î» + 17 = 0 2 Î» 1 , 2 = 8 Â± âˆš 64 4 * 17 2 = 8 Â± 2 i 2 = 4 Â± i. For Î» 1 = 4 + i we have 5 (4 + i ) 2 1 3 (4 + i ) a + bi c + di = ( a + b 2 c ) + ( a + b 2 d ) i ( a c + d ) + ( b c d ) i = . This gives us four equations: a + b 2 c = 0 a + b 2 d = 0 a c + d = 0 b c d = 0 We can now set up the matrix corresponding to the above system of linear equations 1 1 2 1 1 2 1 1 1 1 1 1 â†’ 1 1 2 2 2 2 1 1 1 1 1 1...
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 Fall '08
 ANDREWCHILDS
 Linear Algebra, Algebra, Determinant, Matrices

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