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 February 2nd, 2015, 04:38 AM #1 Newbie   Joined: Feb 2015 From: Brazil Posts: 1 Thanks: 0 self- adjoint operator of R3 3 ) Find a self- adjoint operator of R3 that preserves the space generated by the vector ( 1,2,3). February 4th, 2015, 04:45 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 A linear operator on R[sup]3[/sup] is "self adjoint" if and only if it can be represented as a symmetric 3 by 3 matrix. The space generated by (1, 2, 3) is the set of all vectors of the form (a, 2a, 3a) for any real number a. So you are looking for $\displaystyle \begin{bmatrix}a & b & c \\ b & d & e \\ c & e & f\end{bmatrix}$ such that, for some k, $\displaystyle \begin{bmatrix}a & b & c \\ b & d & e \\ c & e & f\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}= \begin{bmatrix}k \\ 2k\\ 3k\end{bmatrix}$ So we must have a+ 2b+ 3c= k, b+ 2d+ 3e= 2k, and c+ 2e+ 3f= 3k. There are an infinite number of solutions. You want to find one. Tags adjoint, operator Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post maya94 Linear Algebra 2 May 25th, 2014 10:16 AM Sandra93 Linear Algebra 1 May 9th, 2014 03:54 AM bingliantech Linear Algebra 0 May 1st, 2014 05:54 PM tinynerdi Linear Algebra 6 May 10th, 2010 11:03 AM rebecca Linear Algebra 1 January 29th, 2010 06:31 AM

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