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Math Help - Matrices: ABA - BAB = I ?

  1. #1
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    Matrices: ABA - BAB = I ?

    Let A, B be n \times n matrices over \mathbb{C}. Is it possible that ABA - BAB = I, where I is the n \times n identity matrix?

    My suspicion is that the answer is no, though I'm not sure how to go about showing it. Any hints?
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  2. #2
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    Re: Matrices: ABA - BAB = I ?

    Consider A= \begin{bmatrix}\frac{1+\sqrt{5}}{2} & 0 \\ 0 & \frac{1- \sqrt{5}}{2}\end{bmatrix} and B= \begin{bmatrix}\frac{-1+\sqrt{5}}{2} & 0 \\ 0  & \frac{-1-\sqrt{5}}{2}\end{bmatrix}.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Matrices: ABA - BAB = I ?

    Another way: if for all n we choose A=xI (escalar matrix) and B=I we obtain ABA-BAB=I\Leftrightarrow (x^2-x-1)I=0\Leftrightarrow x^2-x-1=0 whose solutions are (1\pm \sqrt{5})/2 . So, for all n we have at least one solution: A=\Phi I,\;B=I where \Phi is the golden ratio.
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