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  1. #1
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    diagonalizable

    hi,

    what does it mean to say that a matrix A is diagonalizable iff R^n has a basis consisting of eigenvectors for A?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If B=\left\{{e_1,\ldots,e_n}\right\} is such a basis then, Ae_1=\lambda_1,\ldots,Ae_n=\lambda_ne_n

    If we denote

    P=\begin{bmatrix}e_1\hdots e_n\end{bmatrix}\;,\quad D=\begin{bmatrix} \lambda_1 & 0 & \ldots & 0\\ 0 & \lambda_2 & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & \lambda_n\end{bmatrix}

    then

    AP=A\begin{bmatrix}e_1 \hdots e_n\end{bmatrix}=\begin{bmatrix}Ae_1 \hdots Ae_n\end{bmatrix}=\begin{bmatrix}\lambda_1e_1 \hdots \lambda_ne_n\end{bmatrix}=\\D\begin{bmatrix}e_1 \hdots e_n\end{bmatrix}=DP

    This means P^{-1}AP=D i.e. A is similar to a diagonal matrix (definition of diagonalizable matrix).

    Try to prove the reciprocal statement, if a is diagonalizable then ...

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    Fernando Revilla
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  3. #3
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    so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?
    If A is similar to a diagonal matrix D, (that is there exists P invertible such that P^{-1}AP=D) try to prove that the columns of P determine a basis of eigenvectors)

    Hint: P^{-1}AP=D is equivalent to AP=PD.

    Regards.

    Fernando Revilla
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  5. #5
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    Quote Originally Posted by alexandrabel90 View Post
    so in this case, instead of B consisting of eigenvectors, B consists of elementary matrices?
    Assuming you mean "B" to be the basis, then, no, a basis is a set of vectors, not a set of matrices. I have no idea where you saw anything related to "elementary matrices" in what FenandoRevilla wrote!
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