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Jordan Forms Problem

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    See Attachment.

    3. The attempt at a solution
    I need an efficient way to find the det(A) so I can find the eigenvalues together with the trace or will use Cayley-Hamilton Thm. I can find the algebraic multiplicities but cannot find the geometric ones. If a matrix is diagonalisable; then AM=GM?
     

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  2. jcsd
  3. Nov 12, 2011 #2

    HallsofIvy

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    You do NOT need to find the determimant of that matrix. You are given one eigenvector and asked to show that it is an eigenector and find the corresponding eigenvalue. Solve the equation
    [tex]\begin{bmatrix} 7 & -1 & 3 & 1 \\ -2 & 6 & -3 & -1 \\ -4 & 2 & -1 & -2 \\4 & -2 & 6 & 7 \end{bmatrix} \begin{bmatrix}1 \\ 2 \\ 0 \\ 0\end{bmatrix}= \lambda \begin{bmatrix}1 \\ 2 \\ 0 \\ 0\end{bmatrix}[/tex]
    for [itex]\lambda[/itex].

    You should then be able to reduce the matrix to 3 by 3 (or smaller) to find the remaining eigenvalues and eigenvectors. An n by n matrix is diagonalizable if and only if it has n independent eigenvectors.
     
  4. Nov 12, 2011 #3
    How would I find the geometric multiplicities? I can do the rest of the question
     
  5. Nov 12, 2011 #4

    I like Serena

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    Assuming you already found λ for the eigenvector.

    Did you write out A-λI?

    You should be able to find other eigenvector solutions easily.
    That is, solutions of (A-λI)v=0.

    This will give you the geometric multiplicity of this λ.

    You should also find that the remaining eigenvalue is easy to deduce as well.
     
  6. Nov 12, 2011 #5
    Ok, thanks. Really should have remembered that. So, the dim{Ker(A-λI)}=Geometric Multiplicity?
     
  7. Nov 13, 2011 #6

    I like Serena

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  8. Nov 13, 2011 #7
    Thanks again.
     
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