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  1. #1
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    matrice 2

    (1)IF A and B are square matrices such that $\displaystyle AB=\omega A$ and $\displaystyle BA=\phi B $ where $\displaystyle \phi $ and $\displaystyle \omega $ are non - zero scalars , prove that $\displaystyle A^2=\phi A $and $\displaystyle B^2=\omega B . $

    (2) If A is a 2x2 matrix such that $\displaystyle A^2=\omega A$ where $\displaystyle \omega$ is a non-zero scalar , prove that A is either non-singular or $\displaystyle A=\phi I $ where I is an identity .


    Thanks for helping me out .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    (1)IF A and B are square matrices such that $\displaystyle AB=\omega A$ and $\displaystyle BA=\phi B $ where $\displaystyle \phi $ and $\displaystyle \omega $ are non - zero scalars , prove that $\displaystyle A^2=\phi A $and $\displaystyle B^2=\omega B . $
    If $\displaystyle AB=\omega A$ then $\displaystyle A = \omega^{-1}AB$, so $\displaystyle A^2 = (\omega^{-1}AB)A =\omega^{-1}A(BA) =\omega^{-1}A(\phi B) = \ldots$.

    Quote Originally Posted by thereddevils View Post
    (2) If A is a 2x2 matrix such that $\displaystyle A^2=\omega A$ where $\displaystyle \omega$ is a non-zero scalar , prove that A is either non-singular or $\displaystyle A=\phi I $ where I is an identity.
    I think you mean "singular", not "non-singular". If A is non-singular then you can multiply both sides of the equation $\displaystyle A^2=\omega A$ by $\displaystyle A^{-1}$ and conclude that $\displaystyle A = \omega I$.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    If $\displaystyle AB=\omega A$ then $\displaystyle A = \omega^{-1}AB$, so $\displaystyle A^2 = (\omega^{-1}AB)A =\omega^{-1}A(BA) =\omega^{-1}A(\phi B) = \ldots$.


    I think you mean "singular", not "non-singular". If A is non-singular then you can multiply both sides of the equation $\displaystyle A^2=\omega A$ by $\displaystyle A^{-1}$ and conclude that $\displaystyle A = \omega I$.
    Thanks Opalg, but how can i continue from here $\displaystyle \omega^{-1}A(\phi B) = \ldots$ , how to get rid of some variables here to get the required result .
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  4. #4
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    Quote Originally Posted by thereddevils View Post
    Thanks Opalg, but how can i continue from here $\displaystyle \omega^{-1}A(\phi B) = \ldots$ , how to get rid of some variables here to get the required result .
    I thought you'd be able to manage the rest of it.

    $\displaystyle \omega^{-1}A(\phi B) = \phi(\omega^{-1}AB) = \phi A$.
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  5. #5
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    Quote Originally Posted by Opalg View Post
    I thought you'd be able to manage the rest of it.

    $\displaystyle \omega^{-1}A(\phi B) = \phi(\omega^{-1}AB) = \phi A$.

    gosh , sorry to disappoint u ..
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