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Find eingenvalues and eigenvectors of an order n matrix

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Being [itex]T\in L(\mathbb{R}^n)[/itex] a linear operador defined by [itex]T(x_1, ... ,x_n )=(x_1+...+x_n,...,x_1+...+x_n )[/itex], find all eigenvalues and eigenvectors of T.

    2. Relevant equations

    [itex]det(T-\lambda I)=0, Ax=\lambda x[/itex]

    3. The attempt at a solution

    By checking n=1,2,3,4 I guess the answer is:

    λ=n, x=(1,1,1)
    λ=0 (multiplicity n-1), x such as , [itex]\forall k \in \{1,...,(n-1)\}[/itex], [itex]x_k=1[/itex], [itex]x_n=-1[/itex] and [itex]x_i=0[/itex] in all other positions. For instance, for n=4, we have (1,0,0,-1), (0,1,0,-1), (0,0,1,-1).

    But how do I prove it for the general case? I'm trying induction, but I think I'm missing something...

    Thanks in advance! :)
     
    Last edited by a moderator: Jun 19, 2011
  2. jcsd
  3. Jun 19, 2011 #2

    HallsofIvy

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    This operator replaces every vector with the vector having each component the sum of the original vectors components. The crucial point is that the image of the vector space under this linear transformation is the space of all vectors of the form (x, x, x, ..., x) with all components the same. That is one dimensional and has {(1, 1, 1, ..., 1)} as a basis. In particular, it maps (1, 1, 1, ..., 1) into the vector (n, n, ..., n) so, as you say, one eigenvalue is n and a corresponding eigenvector is (1, 1, ..., 1).
    Since this maps all of Rn into a single line, there is a second, obvious eigenvalue with n-1 dimensional "eigenspace".
     
    Last edited: Jun 19, 2011
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