My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum

LinkBack Thread Tools Display Modes
February 2nd, 2015, 04:38 AM   #1
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).
Enio Mouzinho is offline  
February 4th, 2015, 04:45 AM   #2
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,240
Thanks: 885

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.
Country Boy is offline  

  My Math Forum > College Math Forum > Linear Algebra

adjoint, operator

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
self-adjoint transformation maya94 Linear Algebra 2 May 25th, 2014 10:16 AM
self-adjoint operator Sandra93 Linear Algebra 1 May 9th, 2014 03:54 AM
what is the difference between self-adjoint operator and hermite matrix? bingliantech Linear Algebra 0 May 1st, 2014 05:54 PM
adjoint of a linear operator tinynerdi Linear Algebra 6 May 10th, 2010 11:03 AM
Adjoint matrix rebecca Linear Algebra 1 January 29th, 2010 06:31 AM

Copyright © 2018 My Math Forum. All rights reserved.