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A = bdb⁻¹

  1. May 25, 2014 #1
    Given a matrix A, is possible to rewrite A like:

    ##A = B D B^{-1}##

    ##
    \begin{bmatrix}
    a_{11} & a_{12} \\
    a_{21} & a_{22} \\
    \end{bmatrix}
    =
    \begin{bmatrix}
    ?_{11} & ?_{12} \\
    ?_{21} & ?_{22} \\
    \end{bmatrix}

    \begin{bmatrix}
    \lambda_{1} & 0 \\
    0 & \lambda_{2} \\
    \end{bmatrix}

    \begin{bmatrix}
    ?_{11} & ?_{12} \\
    ?_{21} & ?_{22} \\
    \end{bmatrix}^{-1}
    ##

    (if A is diagonalizable)

    Being ##\lambda_i## the i-th root of the characterisc polynomial of A.

    But, what is the definition of the matrix B in terms of A?
     
  2. jcsd
  3. May 25, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I believe you are asking if every matrix is "diagonalizable". The answer to that is "not every matrix"! A matrix is diagonalizable if and only if it has a "complete set" of eigenvectors. That is, an n by n matrix is diagonalizable if and only if it has a set of n independent eigenvectors. Since eigenvectors corresponding to distinct eigenvalues are always independent, if all eigenvalues of a matrix are distinct then it is diagonalizable. But a matrix with repeated eigenvalues may still be diagonalizable.

    If A has n independent eigenvectors, then we can construct the matrix B having the eigenvectors of A as columns. Since the eigenvectors are independent, B is invertible and then we have [itex]A= BDB^{-1}[/itex].

    The matrix [itex]\begin{bmatrix}8 & -3 \\ 10 & -3\end{bmatrix}[/itex] has eigenvalues 2 and 3. Eigenvectors corresponding to eigenvalue 2 are multiples of [itex]\begin{bmatrix}1 \\ 2\end{bmatrix}[/itex] and eigenvectors corresponding to eigenvalue 3 are multiples of [itex]\begin{bmatrix}3 \\ 5 \end{bmatrix}[/itex]. So if we let [itex]B= \begin{bmatrix}1 & 3 \\ 2 & 5\end{bmatrix}[/itex], we have [itex]B^{-1}= \begin{bmatrix}-5 & 3 \\ 2 & -1\end{bmatrix}[/itex].

    And then [tex]BDB^{-1}= \begin{bmatrix}1 & 3 \\ 2 & 5\end{bmatrix}\begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}-5 & 3 \\ 2 & -1\end{bmatrix}= \begin{bmatrix}8 & -3 \\ -10 & -3\end{bmatrix}[/tex]

    But, again, not every matrix is diagonalizable. The matrix [itex]\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}[/itex] has 1 as a double eigenvalue but the only eigenvectors are the multiples of [itex]\begin{bmatrix}1 \\ 0 \end{bmatrix}[/itex].
     
    Last edited by a moderator: Jun 1, 2014
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