Solution06_121 - Solutions to PS 6 (Math 121) J. Scholtz...

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Solutions to PS 6 (Math 121) J. Scholtz November 15, 2006 Question 1: 5.1/1 The answers are at the back of the book. Question 2: 5.1/3b Find the eigenvalues, corresponing eigenvectors, if possible the eigenbasis and in such case also the diagonal matrix that is equivalent to the following matrix: A = 0 - 2 - 3 - 1 1 - 1 2 2 5 By evaluating det( A - tI ) = - t 3 +6 t 2 - 11 t +6 = - ( t - 1)( t - 2)( t - 3), we get that the eigenvalues are 3 , 2 , 1. Note that since we have three distinct eigenvalues, of which each must have an eigenspace of dimension at least 1, we now that the matrix is diagonalizable with each eigenspace having dimension exactly 1. Then by finding the nullspaces for A - λI we can identify the eigenspaces: E 1 = span { ( - 1 , - 1 , 1) } , E 2 = span { ( - 1 , 1 , 0) } and E 3 = span { ( - 1 , 0 , 1) } . Therefore the diagonal matrix is: D = 1 0 0 0 2 0 0 0 3 and the change of basis matrix is: Q = - 1 - 1 - 1 - 1 1 0 1 0 1 Question 3: 5.1/4e For the following linear operator find the eigenvalues of T and an ordered basis such that [ T ] β is diagonal. let V = P 2 ( R ) and T ( f ( x )) = xf 0 ( x ) + f (2) x + f (3). let us start in the standard basis: T (1) = x + 1, T ( x ) = 3 x + 3, T ( x 2 ) = 2 x 2 + 4 x + 9. This correpsonds to a matrix: A = 1 3 9 1 3 4 0 0 2 And det( A - tI ) = - t ( t - 2)( t - 4). So our eigenvalues are { 0 , 2 , 4 } and the corresponding eigenvectors are obtained by solving for the nullspaces of A - λI : v 1 = ( - 3 , 1 , 0), v 2 = ( - 3 , - 13 , 4) and v 3 = (1 , 1 , 0) and these define the basis in which [ T ] is diagonal.
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This note was uploaded on 07/03/2011 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.

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Solution06_121 - Solutions to PS 6 (Math 121) J. Scholtz...

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