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!

Basic Eigenvalue Proof

  1. Apr 3, 2010 #1
    1. The problem statement, all variables and given/known data
    If v is an eigenvector of A with corresponding eigenvalue [tex]\lambda[/tex] and c is a scalar, show that v is an eigenvector of A-cI with corresponding eigenvalue [tex]\lambda[/tex]-c.


    2. Relevant equations



    3. The attempt at a solution
    I started out thinking that I have to figure out how to go from:
    Ax=[tex]\lambda[/tex]x
    to
    (A-cI)x=([tex]\lambda[/tex]-c)x

    Is this the right start?
     
  2. jcsd
  3. Apr 3, 2010 #2

    phyzguy

    User Avatar
    Science Advisor

    Looks like a good start to me. Keep going.....
     
  4. Apr 3, 2010 #3
    I don't see how I'm supposed to continue. It just seems obvious to me and I can't think of any intermediate steps.
     
  5. Apr 3, 2010 #4

    phyzguy

    User Avatar
    Science Advisor

    Try expanding (A-cI) x.
     
  6. Apr 3, 2010 #5
    Ok I expanded the first step to...
    Ax-(CI)x=[tex]\lambda[/tex]x-cx
     
  7. Apr 3, 2010 #6
    I also know that CI is similar to cx, except CI is a matrix and cx is not. But I'm stuck...
     
  8. Apr 4, 2010 #7

    phyzguy

    User Avatar
    Science Advisor

    I don't see your problem. You showed that [tex] (A-cI)x = (\lambda-c)x[/tex], which is what you were supposed to show.
     
  9. Apr 4, 2010 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, he didn't show that, he simply wrote it and asked how to go from [itex]Ax= \lamda x[/itex] to that.

    maherelharake, From (A- cI)x you get Ax- (cI)x= Ax- c(Ix). Since x is an eigenvector of A corresponding to eigenvalue [itex]\lambda[/itex], [itex]Ax= \lambda x[/itex]. I is the identity operator so Ix= x. Put those together.
     
  10. Apr 4, 2010 #9
    This is what I did step by step. How does it look?
    (A-cI)x=([tex]\lambda[/tex]-c)x
    Ax-(cI)x=[tex]\lambda[/tex]x-cx
    Ax-c(Ix)=[tex]\lambda[/tex]x-cx
    Ax-cx=[tex]\lambda[/tex]x-cx
    Ax=[tex]\lambda[/tex]x
     
  11. Apr 4, 2010 #10
    Is that sufficient?
     
  12. Apr 4, 2010 #11

    phyzguy

    User Avatar
    Science Advisor

    It seems like you are confused with how to go about demonstrating or proving something. You need to start with something you know, then apply known operations until you arrive at what you are trying to prove. Here, the first line of your proof is what you are trying to prove, and you don't yet know it to be true, so if I were grading this proof I would object to the very first line. Actually if you just reverse the order of steps, starting from the bottom and working up, you have pretty well shown what you are trying to prove
     
  13. Apr 4, 2010 #12
    I see what you are saying. So just reverse it?
     
  14. Apr 4, 2010 #13
    Well I guess I will just go with the reverse of what I had...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic Eigenvalue Proof
  1. Eigenvalue proof (Replies: 1)

  2. Eigenvalue proof (Replies: 23)

  3. Eigenvalue Proof (Replies: 5)

  4. Eigenvalues proof (Replies: 7)

  5. Eigenvalue proof (Replies: 6)

Loading...