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Homework Help: Seeking of eigenvalues and eigenvectors of a given matrix

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data

    in seeking of eigenvalues and eigenvectors of a given matrix A, is it permissible first to simplify A by means of some elementary operation? (that is, are the eigenvalues and eigenvector of A invariant with respect to elementary row operation)? (prove it)


    2. Relevant equations

    n/a

    3. The attempt at a solution

    i want to prove it, but before that i want to translated it correctly

    F is a field, v is eigenvector, λ is eigenvalue

    Given A[tex]\in[/tex]Mnxn(F)

    if B is row equivalent to A, then there exist unique λ[tex]\in[/tex]F and v such that
    Av=λv=Bv

    so, is my translation correct?
     
    Last edited: Jul 23, 2010
  2. jcsd
  3. Jul 23, 2010 #2

    HallsofIvy

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    Science Advisor

    No, unfortunately eigenvalues are changed by "row operations" and so "simplifying" a matrix that way does not help.
     
  4. Jul 23, 2010 #3
    yes i know, i got counter example, but i wan to try to prove it systematically,

    if B is row equivalent to A, then there exist unique λF and v such that
    Av=λv=Bv,

    so maybe i can proof by contradiction or something, but is that statement really same as the question?
     
    Last edited: Jul 23, 2010
  5. Jul 23, 2010 #4
    or i simply just give the counterexample? and done proof?
     
  6. Jul 23, 2010 #5
    Unless you are explicitly asked for a proof, a counterexample is enough
     
  7. Jul 23, 2010 #6
    unless it says "for all elementary matrices" then counter example is ok

    but it said "for some elementary matrices", right?

    now i'm still rereading all my lecture notes on logic, help me if you can, with the logic owhoo,
     
  8. Jul 24, 2010 #7
    can i do like this

    Suppose A=(En....E2E3)B and there exist unique v and λ such that Av=λv and Bv=λv

    then, when λ and v is unique then λv is unique which imply Av=Bv,

    when v is unique, Av=Bv => A=B !!!

    contradict the fact that A=(En....E2E3)B => A[tex]\neq[/tex]B

    Conclusion, If A=(En....E2E3)B then Av[tex]\neq[/tex]λv or Bv[tex]\neq[/tex]λv for all v and λ

    i this really correct? i can't tell whether i'm just doing thing to trivial, help T_T
     
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