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Algebraic Multiplicity of Eigenvalues Question?

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\vec{x}[/itex] and [itex]\vec{v}[/itex] be vectors in [itex]\mathbb{R}^3[/itex].
    If A is a matrix such that A[itex]\vec{x}[/itex] gives the projection of [itex]\vec{x}[/itex] onto [itex]\vec{v}[/itex], then what are the eigenvalues of A and what are their algebraic multiplicities?

    2. Relevant equations
    Eigenvalue: A real number λ is an eigenvalue of A if there exists a nonzero vector [itex]\vec{u}[/itex] such that A[itex]\vec{u}[/itex]=λ[itex]\vec{u}[/itex].

    Algebraic Multiplicity: The algebraic multiplicity of an eigenvalue λ of A is determined by the largest integer k>0 such that (x-λ)k divides the characteristic polynomial of A, pA(x).


    3. The attempt at a solution
    I feel like for any λ[itex]\in{\mathbb{R}^3}[/itex] we can find a vector [itex]\vec{x}[/itex] such that A[itex]\vec{x}[/itex]=λ[itex]\vec{x}[/itex] because letting [itex]\vec{x}[/itex]=λ[itex]\vec{v}[/itex] lets us say Aλ[itex]\vec{v}[/itex]=λ[itex]\vec{v}[/itex] since A[itex]\vec{x}[/itex] just projects [itex]\vec{x}[/itex] onto [itex]\vec{v}[/itex] so projecting λ[itex]\vec{v}[/itex] onto [itex]\vec{v}[/itex] obviously just gives λ[itex]\vec{v}[/itex].

    However, if for all [itex]k\in{\mathbb{R}}[/itex], [itex]\vec{u}\in{\mathbb{R}^3}\neq k\vec{v}[/itex] then [itex]\vec{u}[/itex] cannot possibly be an eigenvector. So this means the eigenspace is just all scalar multiples of [itex]\vec{v}[/itex].

    Another thing I think may be significant is that each eigenvalue has only one eigenvector corresponding to it; does this mean the algebraic multiplicity of every eigenvalue is 1? I can't think of a way to find out if (x-λ)k divides the characteristic polynomial of A... is there another way to find out the algebraic multiplicity of an eigenvalue?




    Thanks
     
  2. jcsd
  3. Nov 17, 2012 #2
    What happens when A is applied to any vector orthogonal to v? What is the dimension of the subspace of such vectors?
     
  4. Nov 17, 2012 #3
    If [itex]\vec{p}[/itex] is a vector orthogonal to [itex]\vec{v}[/itex], then A[itex]\vec{p}[/itex]=0[itex]\vec{v}[/itex]=[itex]\vec{0}[/itex].

    The dimension of the subspace of vectors orthogonal to [itex]\vec{v}[/itex] is the plane such that when you apply A to any vector on the plane you get [itex]\vec{0}[/itex]. (I don't know what to call this; I was going to say [itex]\mathbb{R}^2[/itex] because it's a plane but I don't think that's right.)


    So I now see in my original post I left out a whole plane of eigenvectors who all correspond to the eigenvalue 0. So from the set of orthogonal vectors to [itex]\vec{v}[/itex] we can get 2 linearly independent eigenvectors corresponding to eigenvalue 0, and from the set of vectors who are collinear to [itex]\vec{v}[/itex] each eigenvector only corresponds to one eigenvalue. So does that mean eigenvalue 0 has algebraic multiplicity 2 and every other eigenvalue (which makes up all real numbers other than 0) has only one eigenvector and thus algebraic multiplicity 1? Is that how it works?
     
  5. Nov 17, 2012 #4
    You are right about eigenvalue 0. However, I do not understand how you ended up with a whole line for the other eigenvalue. Take any vector u that is a multiple of v. The result of A's action will that same vector u. What is the eigenvalue then?
     
  6. Nov 17, 2012 #5
    A(k[itex]\vec{v}[/itex])=k[itex]\vec{v}[/itex].

    So what I was trying to say is that the only eigenvector that corresponds to the eigenvalue k is k[itex]\vec{v}[/itex]. AKA, there is only one eigenvector for each nonzero eigenvalue.
    Is this right and does it mean the multiplicity for all real numbers other than 0 is 1?

    EDIT: Oh nevermind!!!! A(k[itex]\vec{v}[/itex])=k[itex]\vec{v}[/itex] means the eigenvalue is 1. So we have infinite eigenvectors all corresponding to eigenvalue 1. So does that mean the algebraic multiplicity for λ=1 is infinity?
    The below question still stands...


    Also, I can't find anything in my book that explains why the fact that there exist two linearly independent eigenvectors that both correspond to eigenvalue 0 implies that the algebraic multiplicity of λ=0 is 2? My book only defines the algebraic multiplicity of an eigenvalue of A as the largest number k≥ 0 such that (λ-x)k divides the characteristic polynomial of A . Can you help me understand?

    Thanks so much for your help so far; I feel like I'm almost there!
     
  7. Nov 17, 2012 #6
    k is not an eigenvalue. If you had ## A\vec{v} = k \vec{v} ##, then k would be one. But you really just have ## A \vec{u} = \vec{u} ##. What is the eigenvalue here?

    You have found the geometric multiplicity of an eigenvalue. It is always equal or less than its algebraic multiplicity. But, since dimension of the space is 3, and there is another eigenvalue, the algebraic multiplicity cannot be greater than 2, hence it is exactly 2.
     
  8. Nov 17, 2012 #7
    Ah, so 1's eigenspace is 1-dimensional (because all eigenvectors that correspond to eigenvalue 1 are linearly dependent), and 0's eigenspace is 2-dimensional (because there are two linearly independent eigenvectors on the plane orthogonal to [itex]\vec{v}[/itex]). And so we know
    1≤(algebraic multiplicity of λ=1)
    and
    2≤(algebraic multiplicity of λ=0)

    But we also know that the characteristic polynomial of a 3x3 matrix is a degree-3 polynomial and by the fundamental theorem of algebra has at most 3 roots. So we know the sum of the algebraic multiplicities is less than or equal to 3.

    So λ=1 must have algebraic multiplicity 1 and λ=0 must have algebraic multiplicity 2.


    Alright! thanks so much voko!
     
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