finsol2003 - M346 Final Exam Solutions, December 10, 2003...

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Unformatted text preview: M346 Final Exam Solutions, December 10, 2003 1. In the vector space R 2 [ t ], consider the basis B = { 1 , 1 + t, 1 + t + t 2 } , the basis D = { 1 + t + t 2 , t + t 2 , t 2 } and the vector v = 2 t 2- 1. a) Find [ v ] B and [ v ] D . (That is, find the coordinates of v in the B basis, and the coordinates of v in the D basis.) Since v = 2 t 2- 1 = 2( b 3- b 2 )- b 1 = 2 b 3- 2 b 2- b 1 , [ v ] B = (- 1 ,- 2 , 2) T . Since v = 2 t 2- 1 = 2 d 3- 1 = 2 d 3 + d 2- d 1 , [ v ] D = (- 1 , 1 , 2) T . b) Find the change-of-basis matrices P BD and P DB . b 1 = d 1- d 2 , b 2 = d 1- d 3 and b 3 = d 1 , so P DB = 1 1 1- 1- 1 . Similarly, d 1 = b 3 , d 2 = b 3- b 1 , d 3 = b 3- b 2 , so P BD = - 1- 1 1 1 1 . The results of parts (a) and (b) could also have been obtained by writing down P EB and P ED , taking their inverses to get P BE and P DE , using these to compute [ v ] B and [ v ] D , and taking products of matrices to get P BD and P DB . 2. Consider the operator L : R 2 [ t ] R 2 [ t ] defined by ( L p )( t ) = p ( t +1)- p ( t ). a) Find the matrix of L in the standard basis { 1 , t, t 2 } . Let e 1 = 1, e 2 = t , e 3 = t 2 . We compute L e 1 = 1- 1 = 0, L e 2 = ( t + 1)- t = 1 = e 1 and L e 3 = ( t + 1) 2- t 2 = 2 t + 1 = e 1 + 2 e 2 , so [ L ] E = 1 1 2 . b) Find the matrix of L in the basis B = { 1 , 1 + t, 1 + t + t 2 } (this is the same basis B you saw in problem 1). Either compute [ L ] B = P BE [ L ] E P EB or work directly in the B basis: L b 1 = 0, L b 2 = 1 = b 1 , L b 3 = 2 t + 2 = 2 b 2 . Either way, we have [ L ] B = 1 2 . 1 3. Matrices and eigenvalues: a) Find the eigenvalues and eigenvectors of the matrix 2 1 4 2 1 1 2 . This is block triangular. The eigenvalue of the upper left block is 1 = 2, while those of the lower right block are 2 = 3 and 3 = 1. The corresponding eigenvectors (obtained by solving ( I- A ) x = 0) are b 1 = (1 , , 0) T , b 2 = (5 , 1 , 1) T , b 3 = (3 , 1 ,- 1) T . b) Find a matrix whose eigenvalues are 1 , 3 , 4 and eigenvectors are 1 3 1 2 2 , 1 3 2 1- 2 and 1 3 - 2 2- 1 . [Hint: there is an easy way to compute P- 1 ] Note that the three eigenvectors are orthonormal, so the matrix...
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This note was uploaded on 04/10/2011 for the course M 346 taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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finsol2003 - M346 Final Exam Solutions, December 10, 2003...

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