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Diagonalized 3x3 Matrix

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data

    quick question, if there is a 3x3 matrix which has exactly 3 distinct eigenvalues why must it be diagonalizable?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 18, 2008 #2


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    How is diagolization related to the dimension of the eigenspace? Why would distinct eigenvalues let you satisfy this dimension criteria?
  4. Dec 18, 2008 #3
    well since its a 3x3 matrix, the most eigenvalues it could have is 3? so since there is 3 unique eigenvalues, then it is definitely diagonalizable?
  5. Dec 18, 2008 #4


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    You need two theorems to show this is true and I hinted at them in my post.
  6. Dec 18, 2008 #5


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    Staff Emeritus
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    I think it is better to think in terms of linear transformations- any linear transformation can be represented as a matrix in a given basis: Apply the linear tranformation to each of the basis vectors in turn, the write the result as a linear combination of the basis vectors- the coefficients are the columns of the matrix.

    The three distinct eigenvalues must have 3 independent eigenvectors. Using those eigenvectors as a basis for the vector space, the linear operator is represented by a diagonal matrix with the eigenvalues on the diagonal.

    Equivalently, if A is a matrix with three distinct eigenvalues, B is the matrix having those three eigenvectors as columns, then B-1AB is the diagonal matrix having the eigenvalues on the diagonal.

    By the way, having three independent eigenvectors is a necessary condition for a matrix to be diagonalizable. Having three distinct eigenvalues is not necessary.
  7. Dec 18, 2008 #6
    so what if a 3x3 matrix only has two eigenvalues, does that mean its not able to be diagonalized?
  8. Dec 18, 2008 #7


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    No, reread what Halls wrote. One of the eigenvalues could have a two dimensional eigenspace.
  9. Dec 18, 2008 #8
    Thanks guys, I think its making sense, we will see on finals week lol.
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