1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is this the obvious? (graph Theory)

  1. Dec 12, 2009 #1
    http://www.win.tue.nl/~aeb/srgbk/node10.html [Broken]

    Under the 'For connected graphs all is clear from above'

    I've been asked to prove this for connected graphs, but I don't see how this is so clear?

    Can anyone help me? Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 13, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It says "it is clear from the above" but you have cut off the "above"! What did it say before that theorem?
     
  4. Dec 13, 2009 #3
    Well I googled my problem and found that link after some pages. I'm not sure what it said.

    But it probably was something like the Perron Frobenius theorem.

    From my notes I have (Perron Frobenius Thereom)

    'Let A be in the set of all Schrodinger operators, and suppose a graph G is connected.

    Then eigenvalue(1) is simple (i.e. has multiplicity 1) and its eigenspace is generated by an eigenvector that is strictly positive everywhere.'


    My general question that I was asked to do came from this email he sent me:

    'These graphs are bipartite, meaning that if you change the sign of every other entry in an eigenvector then you get an eigenvector associated with negative the first eigenvalue (prove this!)'
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why is this the obvious? (graph Theory)
  1. Graph Theory (Replies: 5)

  2. Graph theory (Replies: 0)

  3. Graph Theory (Replies: 2)

  4. Graph Theory (Replies: 0)

  5. Graph Theory (Replies: 4)

Loading...