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Eigen value 0 prove

  1. Mar 6, 2009 #1
    prove that a matrix cannot be invertible if it has eigen value of 0.

    not invertible mean that the determinant equals 0.

    x(x-2)=0
    ??
     
  2. jcsd
  3. Mar 6, 2009 #2

    Mark44

    Staff: Mentor

    Where does x(x -2) come in?

    You have an n x n matrix A that has 0 as an eigenvalue. How do you find the eigenvalues of a matrix? How are eigenvalues defined?
     
  4. Mar 6, 2009 #3
    x(x -2) its just an example for a polynomial which is not inertible
    i thought of proving like this

    that if an operator T is not invertible then det T=0 then dim ker differs 0
    (we have a row of zeros) so i have eigen value 0

    correct
    ??
     
    Last edited: Mar 6, 2009
  5. Mar 6, 2009 #4
    "If an equation containing a variable parameter possesses nontrivial solutions only for certain special values of the parameter, these solutions are called eigenfunctions and the special values are called eigenvalues."

    how to use it
    ??
     
  6. Mar 6, 2009 #5

    Office_Shredder

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    Gold Member

    Not the kind of eigenvalue you want (well, sort of is, but let's not complicate things). An eigenvalue [tex] \lambda [/tex] of a matrix A is a number such that there exists v in the vector space, v non-zero, with

    [tex]Av = \lambda v[/tex]
     
  7. Mar 6, 2009 #6
    so if i have eigen value zero
    Av=0

    how does it show that its not invertible
    i know that now v plays the role of kernel A
    but i dont know what is the link between the kernel of an operator and its
    ability to be invertible
    ??
     
  8. Mar 6, 2009 #7

    Office_Shredder

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    A matrix is invertible if it has... an inverse. To have an inverse, it must be 1-1. Think about what having a non-trivial kernel means with regards to this
     
  9. Mar 6, 2009 #8
    1-1 means that for each x i have a y value and vise versa
    i really dont know what is the affect of kernel
    i know how to find it in a matrix
    i know
    dim Im + dim Ker =V

    ??
     
  10. Mar 6, 2009 #9

    statdad

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    Homework Helper

    Remember that in order for a vector [tex] v [/tex] to be an eigenvector, it must be true that [tex] v \ne 0[/tex]. If that is the case, yet [tex] A v = 0 [tex], what can you conclude?
     
  11. Mar 6, 2009 #10
    zero vector is the product of multiplication of the operarator by the kernel vectors

    i dont know
    ??
     
  12. Mar 6, 2009 #11

    statdad

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    Well, remember that if the matrix [tex] A [/tex] is invertible, the solution to

    [tex]
    Ax = b
    [/tex]

    is always given by

    [tex]
    x = A^{-1}b
    [/tex]

    Now, suppose that [tex] v \ne 0 [/tex] is an eigenvector that corresponds to the eigenvalue [tex] \lambda = 0 [/tex]

    Now assume that [tex] A [/tex] is invertible. What does the equation

    [tex]
    Av = \lamba \, v = 0
    [/tex]

    tell you about [tex] v [/tex]? What does this contradict, and what assumption led to that contradiction?
     
  13. Mar 6, 2009 #12
    Can you use the fact that the inverse is the matrix of cofactors divided by the determinant, and that division would be undefined if you had a zero eigenvalue because the determinant is the product of the eigenvalues?
     
  14. Mar 6, 2009 #13
    Consider
    Ax = 0
    A0 = 0

    What is A-1(0)?
     
  15. Mar 7, 2009 #14
    what is the answer
    ??
     
  16. Mar 7, 2009 #15
    why if dim ker differs 0
    then its not invertible
    ??
     
  17. Mar 7, 2009 #16

    Mark44

    Staff: Mentor

    Let's get organized. You're trying to prove this statement, which is a paraphrase of your first post in this thread:

    If matrix A has 0 as an eigenvalue, A does not have an inverse.​

    If A has 0 as an eigenvalue, Ax = 0x = 0, for some nonzero vector x (the eigenvector associated with eigenvalue 0).

    So Ax = 0, where x is not the zero vector.
    From this equation, is it possible to say either that A definitely has an inverse or that A definitely doesn't have an inverse?
     
  18. Mar 7, 2009 #17
    i can say that definitely we have a kernel vector in A
    but i dont know how it affects the invertiblity of A
    ??
     
  19. Mar 7, 2009 #18

    Mark44

    Staff: Mentor

    There is always at least one vector in ker(A), regardless of whether A has an inverse.
     
  20. Mar 7, 2009 #19
    yes 0,0,0,
     
  21. Mar 7, 2009 #20

    HallsofIvy

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    This has already been said, but I'll try again. If Ax= 0 and A has an inverse, what do you get if you multiply both sides of that equation by A-1? What does that tell you about the kernel of A? How is that a contradiction?
     
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