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Thread: Matrices: ABA - BAB = I ?

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

    Let $\displaystyle A, B$ be $\displaystyle n \times n$ matrices over $\displaystyle \mathbb{C}$. Is it possible that $\displaystyle ABA - BAB = I$, where $\displaystyle I$ is the $\displaystyle n \times n$ 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 $\displaystyle A= \begin{bmatrix}\frac{1+\sqrt{5}}{2} & 0 \\ 0 & \frac{1- \sqrt{5}}{2}\end{bmatrix}$ and $\displaystyle B= \begin{bmatrix}\frac{-1+\sqrt{5}}{2} & 0 \\ 0 & \frac{-1-\sqrt{5}}{2}\end{bmatrix}$.
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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 $\displaystyle n$ we choose $\displaystyle A=xI$ (escalar matrix) and $\displaystyle B=I$ we obtain $\displaystyle ABA-BAB=I\Leftrightarrow (x^2-x-1)I=0\Leftrightarrow x^2-x-1=0$ whose solutions are $\displaystyle (1\pm \sqrt{5})/2$ . So, for all $\displaystyle n$ we have at least one solution: $\displaystyle A=\Phi I,\;B=I$ where $\displaystyle \Phi$ is the golden ratio.
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