My Math Forum self- adjoint operator of R3

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 February 2nd, 2015, 05:38 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, 05:45 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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.

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