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Proof involving char. poly., a basis, and similarity to an upper-triangular matrix

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Let A be an n x n matrix and suppose that the characteristic polynomial of A is
    p(t) = ([tex]\lambda[/tex]-t)[tex]^{n}[/tex] for some fixed [tex]\lambda[/tex] [tex]\in[/tex] R. Show that A is similar to an upper-triangular matrix. Do this by finding a basis with the following property: {x1,...,xn} is a basis of R[tex]^{n}[/tex] such that Axk [tex]\in[/tex] span{x1,...,xk} for k = 1,2,...,n. You may find the Cayley-Hamilton Theorem helpful.


    2. Relevant equations
    p(t) = det(A-t*I) = ([tex]\lambda[/tex]-t)[tex]^{n}[/tex]
    p(A) = 0 (Cayley-Hamilton Theorem)


    3. The attempt at a solution
    I have already done the following problem:
    Let A be an n x n matrix and let {x1,...,xn} be a basis of R[tex]^{n}[/tex]. Show that the matrix of the linear transformation associated to A with respect to this basis is upper-triangular if and only if Axk [tex]\in[/tex] span{x1,...,xk} for k = 1,2,...,n.

    I think the matrix [x1,...,xn] must itself be upper-triangular. Such a basis would not be difficult to find, however I don't understand how to connect this with the Cayley-Hamilton Theorem or the fact that [tex]\lambda[/tex] is an eigenvalue of A with multiplicity n. I also don't have any good ideas of how to go about the proof in general.

    Any help would be greatly appreciated!
     
  2. jcsd
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