1. Limited time only! Sign up for a free 30min personal 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!

A linear algebra proof

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that A has c=0 as an eigenvalue if and only if A is non invertible.


    2. Relevant equations



    3. The attempt at a solution
    Let A be a square matrix (mxm). Then Av=(a1,1v1+...+am,1vm)+...+(a1,mv1+...+am,mvm).
    Since an identity matrix is square, and Icv is an mxm matrix,
    we have (a1,1v1+...+a1,mv1)+..+(am,1vm+...+am,mvm)
    For (a1,1+...+a1,m)v1+...+(am,1+...+am,m)vm=Icv,
    (ak,1+...+ak,m) must=c. But if the matrix is mxn, n>m
    then we have (ak,1+...+ak,m+...+ak,n) which is>(ak,1+...+ak,m).
    Unless all rows after m are 0 (making the matrix back to mxm)
    all (ak,1+...+ak,n) must be zero along with c so that
    Av=0=Icv=I(0)v=0.
     
  2. jcsd
  3. Jun 11, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You don't need to write out any components of anything. Define what it means for a matrix to be noninvertible in terms of the kernel.
     
  4. Jun 11, 2009 #3
    Hey Dick, hows it going? :smile: Anyway, in terms of the kernel, a matrix is noninvertible
    if the transformation is not injective, so the kernel has elements besides zero.
    So it is not just 0 that gets mapped to 0. Within the kernel are those vectors
    in rows that become 0 after row reducing. So letting v be a 1xn vector,
    and A having more rows than v but same amount of columns as v has rows,
    we multiply and row reduce to see that the excessive rows become 0.
     
    Last edited: Jun 11, 2009
  5. Jun 11, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Great, how about with you? Ok, so if the matrix is noninvertible there is a nonzero vector v such that Av=0. What does that tell you about eigenvalues of A?
     
  6. Jun 11, 2009 #5
    If Av=0 and v=/=0 then A is the zero transformation of v if v is in the kernel. So for Av to =0
    knowing v is not 0, v must be multiplied by 0 to get to 0.
    Since Iv=v=/=0, and Av=0*v,
    0*v=Icv=cIv=0*Iv=0*v
     
    Last edited: Jun 11, 2009
  7. Jun 11, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's all in there, but it doesn't clearly answer my question. I think you mean to say simply, if Av=0 and v is nonzero then 0 is an eigenvalue because _______. Try and make a clear statement of that 'because'. Start with the definition of an eigenvalue and say why 0 is one.
     
  8. Jun 11, 2009 #7
    v is not zero, so the eigenvalue must=0 Av=cv=0*v=0
     
  9. Jun 11, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, so you've got "if A is noninvertible then 0 is an eigenvalue". Now do the converse. "if 0 is an eigenvalue then A is noninvertible".
     
  10. Jun 11, 2009 #9
    If 0 is an eigenvalue, then there is some nonzero v in the kernel of A such that Av=0.
     
  11. Jun 11, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I would say "If 0 is an eigenvalue then there is a nonzero vector v such that Av=0*v=0. So v is in the kernel of A. Since the kernel is nontrivial A is noninvertible." Compare that with your version, which doesn't even contain the word 'noninvertible'. A proof something is noninvertible should probably contain the word noninvertible.
     
  12. Jun 11, 2009 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Good advice!
     
  13. Jun 11, 2009 #12
    That makes sense :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A linear algebra proof
  1. Linear Algebra Proof (Replies: 8)

  2. Linear Algebra proof (Replies: 21)

Loading...