Eigenvalue, Eigenvector and Eigenspace

  1. Let's say my eigenvalue λ=-1 and we assume eigenvector of zero are non-eigenvector.
    An eigenspace is mathematically represented as Eλ = N(λ.In-A) which essentially states, in natural language, the eigenspace is the nullspace of a matrix.
    N(λ.In-A) is a matrix.

    Would it then be valid to say that the eigenspace, Eλ, whose eigenvalue, λ=-1, is the nullspace of the matrix, N(λ.In-A), is equivalent to the the vector , v, where
    Av = 0.

    If v is the nullspace of the matrix A then Av = 0, and similarly, if Eλ is the nullspace of a matrix, N(λ.In-A), then, it must equally be true that
    [ N(λ.In-A) ] [Eλ=-1] = 0
    Last edited: May 10, 2014
  2. jcsd
  3. Simon Bridge

    Simon Bridge 14,458
    Science Advisor
    Homework Helper
    Gold Member

    You mean you are trying to construct an eigenspace from a single vector?
  4. lurflurf

    lurflurf 2,325
    Homework Helper

    It is not clear exactly what you mean by
    [ N(λ.In-A) ] [Eλ=-1] = 0
    Certainly some interpretations would make the statement true, but
    [ N(λ.In-A) ] =[Eλ=-1]
    is also true (some interpretations)
  5. HallsofIvy

    HallsofIvy 40,241
    Staff Emeritus
    Science Advisor

    "We assume eigenvector of zero are non-eigenvector"? I don't know what this means. Do you mean that you are assuming that 0 is not an eigenvector or are you asserting that an "eigenvector" must not be the 0 vector?

    No. First, it is a set of vectors, not a single vector. Second, it is the set of vectors such that (λ.In-A)v= 0, or equivalently, λ.Inv= Av, as you said, not "Av= 0".

    I cant make sense out of that. You have, as part of your hypothesis, "If v is (in) the nullspace of the matrix A" but you conclusion has no "v" in it.
  6. You probably can't because it doesn't make sense. It's still a new topic to me so I'd figure the fastest way to refine my understanding is to state what I know and have it corrected where the understanding is flawed.

    I'm stating that a zero vector is not within the consideration of an eigenvector as I've learnt.

    Let's see:

    λ is an eigenvalue of the eigenvetor V IFF det(λIn-A) = 0

    Say we have a matrix [1 2;4 3] and det((λIn-A) = 0 gives
    λ1=5 and λ2 = -1

    1) Does this then states that λ1=5 and λ2 = -1 are eigenvalues corresponding to the eigenvector V?

    2) Eλ12=Null(λIn-A) is interpreted as the set of vector with eigenvalues of λ1 and λ2 that maps
    the matrix (λIn-A) to the zero vector. True?

    3) where λ1=-1:

    I have Eλ1=N(λIn-A) = N(-1[1 0; 0 1] - [1 2; 4 3])
    = N([-2 -2; -4 -4])

    4) (I shall take only λ1 as an example. Since Eλ1 maps the matrix [-2 -2; -4 -4] to the zero vector then could it then be written
    as AV=0
    where A = [1 2; 4 3] and V is the set of vector V = {v1, v2,....vn} that maps the matrix A to the zero vector?

    If I could, then, matrix A after RREF(is it a necessarily condition to perform RREF on matrix A? what is the implication if I perform REF instead of RREF?)
    becomes [ 1 1; 0 0 ] and V = {v1, v2}
    and so,
    [ 1 1; 0 0] [v1;v2] = [0;0]
  7. Mark44

    Staff: Mentor

    By definition, an eigenvector is nonzero.
    These are eigenvalues of that matrix. Each eigenvalue has its own eigenvector. For the eigenvalue λ1 = 5, an eigenvector associated with this eigenvalue is any nonzero solution x of the matrix equation Ax = 5x. Equivalently, x is a nonzero solution of (A - 5I)x = 0.
    Not sure. This is a pretty cumbersome way to say it.
    For a given eigenvalue λ of a matrix A, x is an eigenvector if it is a nonzero solution of the equation Ax = λx.
    This is equivalent to (A - λI)x = 0, so x is any nonzero vector in the nullspace of A - λI.
    No. You're not looking for solutions of Ax = 0 - you're looking for solutions of (A - λI)x = 0.
  8. HallsofIvy

    HallsofIvy 40,241
    Staff Emeritus
    Science Advisor

    I'm going to jump in here- Yes, many textbooks define an eigenvector to be non-zero. But many others define "Eigenvalue" by "[itex]\lambda[/itex] is an eigenvalue for linear operator A if there exist a non-zero vector, v, such that [itex]Av= \lambda v[/itex]" and then define an Eigenvector, corresponding to eigenvalue [itex]\lambda[/itex], to be any vector, v, such that [itex]Av= \lambda v[/itex], which includes the 0 vector.

    Personally, I prefer that because it allows us to say "the set of all eigenvectors, corresponding
    to eigenvalue [itex]\lambda[/itex], is a subspace" rather than having to say "the set of all eigenvectors, corresponding to eigenvalue [itex]\lambda[/itex], together with the 0 vector, is a subspace"
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