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Math Help - exams very soon -help-Diagonalization

  1. #1
    n22
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    exams very soon -help-Diagonalization

    Hi people,
    Can someone tell me when a matrix actually becomes diagonizable??
    and when eigenvalue is considered distinct and not distinct?
    Thanks.
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  2. #2
    MHF Contributor
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    Re: exams very soon -help-Diagonalization

    Hey n22.

    Hint: Given a characteristic equation, how can you obtain the eigen-values? (Also if you have repeated eigen-values what does this mean for the roots of the characteristic polynomial?)
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  3. #3
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    Re: exams very soon -help-Diagonalization

    Quote Originally Posted by n22 View Post
    Hi people,
    Can someone tell me when a matrix actually becomes diagonizable??
    and when eigenvalue is considered distinct and not distinct?
    Thanks.
    It doesn't make sense to talk about an eigenvalue being "distinct". Two or more eigenvalues are "distinct" if they are different! That is, after all, what "distinct" means.

    An n by n matrix is diagonalizable if and only it has n independent eigenvectors. It can be shown that "distinct" (i. e "different") eigenvalues have eigenvectors that are independent eigenvectors so if an n by n matrix has n distinct eigenvalues, it has n independent eigenvectors and so is diagonalizable. The converse is NOT true. If a matrix has multiple eigenvalues it may or may not be diagonalizable.

    For example if A= \begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}
    B= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix} and
    C= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{bmatrix}
    Then each of A, B, and C have "3" as a triple eigenvalue. A has three independent eigenvectors, B has two independent eigenvectors, and C has only one independent eigenvector.
    A can be "diagonalize" (clearly it is diagonal) but B and C cannot.
    Thanks from n22
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