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Distinct Eigenvalues

  1. May 9, 2010 #1
    Can zero be a distinct eigernvalue?
  2. jcsd
  3. May 9, 2010 #2
    Imagine you get to a characteristic equation

    [tex]\lambda (\lambda-1)(\lambda -2)=0[/tex]

    during your calculations. Perfectly valid. Now what would your eigenvalues be? :wink:
  4. May 9, 2010 #3
    Yes the eigenvalues are 0 1 and 2 but is 0 distinct?
  5. May 9, 2010 #4
    Yes, it is no more special than having 1 or 2. All it means is that your matrix is singular. What we can't have though is a 0 eigenvector (for reasons that I can't recall right now).
  6. May 9, 2010 #5
    This is what I thought, problem is that it messes with a proof I'm trying to complete.

    I have to prove the following;

    Let V be a finite dimensional vector space and T: V to V be a linear transformation. Let a1,...,ak be distinct eigenvalues of T and let v1,..,vk be corresponding eigenvectors. Prove that {v1,...,vk} is linearly independent.

    If it was the case that 0 cannot be a distinct eigenvalue then I can complete the proof but since this isn't the case I'm a liitle stuck! Could anyone possible help me out here?
  7. May 9, 2010 #6


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    I don't understand how one number can be 'distinct', but anyway: 0 is perfectly allowed as eigenvalue, i.e. one of the a_i may or may not equal 0. Please write out explicitly what proof you had in mind and where you're stuck, then we can help you!
  8. May 9, 2010 #7
    Well basically the proof (as far as I can see) is asking to show that the sum from i=1 to k of alpha(i)v(i) =0 implies that alpha(i)=0 for all i, alpha being an element of the field F.
    Now taking x element of V, x = sum from i=1 to k of alpha(i)v(i)
    then T(x) = sum from i=1 to k of alpha(i)T(v(i)) = sum from i=1 to k of alpha(i)a(i)v(i)
    Now for this sum to equal zero then either aplha(i)=0 (as required) OR a(i)v(i)=0, if it was the case that a(i) could not be equal to zero then the prrof would be complete, but since there could be one case where a(i)=0 I am stuck and don't know how to move on from here.
    Sorry I haven't Latexed anything, I don't know how!
  9. May 9, 2010 #8


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    Hmm, this doesn't look right.
    Yes, so far so good. However:
    How so?! No, this is not a correct conclusion. You are basically saying if a+b=0 then a=0 and b=0.

    A more fruitful approach: by contradiction, assume that {v1,...,vk} is dependent. Let [itex]j\leq k[/tex] be the smallest index such that [itex]v_j\in\mbox{span}(v_1,...v_{j-1})[/itex], i.e. j is minimal with the property of having a non-trivial linear combination
    [tex]\alpha_1v_1+...+\alpha_j v_j=0[/tex].
    Now apply T to this equation, and think some more.
  10. May 10, 2010 #9


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    "Distinct" numbers just means different numbers. If a and b are eigen values of operator T and [itex]a\ne b[/itex] then they are "distinct" eigenvalues. If they happen to be 0 and 1, then, since they are different, they are "distinct". That's why Landau said "I don't understand how one number can be 'distinct'". The word "distinct" doesn't apply to a single number, it applies to sets of numbers- the numbers in the set are "distinct" if and only if they are all different.

    phyzmatic, 0 can be an eigenvector. A number, [itex]\lambda[/itex] is defined to be an eigenvalue of operator T if and only if there exist a non-zero vector v such that [itex]Tv= \lambda v[/itex] but once we have that any vector, including 0, such that [itex]Tv= \lambda v[/itex] is an eigenvector of T.

    We need to have 0 an eigenvector in order to say "the set of all eigenvectors of T, corresponding to eigenvalue [itex]\lambda[/itex], is a subspace".
  11. May 10, 2010 #10


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    That's just a matter of definition. A lot of people exclude 0 from being an eigenvector. They then say "the set of all eigenvectors, together with the zero vector, is a subspace". Other authors, like Axler, do the same as you. It doesn't matter, as long as you be clear on your definition.
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