
Linear Algebra Linear Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. 

Tags 
adjoint, operator 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
selfadjoint transformation  maya94  Linear Algebra  2  May 25th, 2014 11:16 AM 
selfadjoint operator  Sandra93  Linear Algebra  1  May 9th, 2014 04:54 AM 
what is the difference between selfadjoint operator and hermite matrix?  bingliantech  Linear Algebra  0  May 1st, 2014 06:54 PM 
adjoint of a linear operator  tinynerdi  Linear Algebra  6  May 10th, 2010 12:03 PM 
Adjoint matrix  rebecca  Linear Algebra  1  January 29th, 2010 07:31 AM 