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Non-linearly independent eigenvectors

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Let B = 1 0
    6 -1
    Be a square matrix. Find a non-singular matrix P such that P-1BP = D, where D is a diagonal matrix and show that P-1BP = D.

    2. Relevant equations

    det(lambdaI - A) = 0

    3. The attempt at a solution

    Ok, this might look like a simple problem...but I get non-linearly independent eigenvectors, even although I have 2 distinct eigenvalues?

    Noting that it is a triangular matrix, the diagonal entries thus correspond to its eigenvalues. So, the matrix has lambda = 1 and lambda = -1 as its eigenvalues.

    For lambda = 1, I obtain the basis vector (1/3, 1).

    For lambda = -1, I obtain the basis vectors (1, 0) and (0, 1)...

    Thus, I have non-linearly independent eigenvectors, since (1/3, 1) can be written as 1/3(1, 0) + (0, 1)...

    ???
     
  2. jcsd
  3. Aug 22, 2011 #2
    Re: diagonalization

    i got only (0,1) and (1/3,1) as the eigenvectors. (1,0) isnt an eigenvector.
     
  4. Aug 22, 2011 #3
    Re: diagonalization

    Ok. I see it now...flip, just needed some revising on finding solution spaces...I get the same answer now too...thanks...
     
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