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Equal determinants [entire topic moved to new forum]

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Let \( A, B \in \mathbb{R}^{n \times n}\)  that are diagonizable in \( \mathbb{R}\) . If \( \det (A^2+B^2)=0\)  and \( AB=BA \)  , then prove that \( \det A = \det B =0\) .

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Accepted Answer

  • Replied by Papapetros Vaggelis on Wednesday, November 04 2015, 02:51 PM
    The matrices \(\displaystyle{A\,,B\in\mathbb{M}_{n}(\mathbb{R})}\) are simultaneously diagonizable, that is there exists an invertible matrix

    \(\displaystyle{C\in\mathbb{M}_{n}(\mathbb{R})}\) and diagonizable matrices \(\displaystyle{P\,,Q\in\mathbb{M}_{n}(\mathbb{R})}\)

    such that \(\displaystyle{A=C\,P\,C^{-1}\,\,,B=C\,Q\,C^{-1}}\) . Therefore,

    $$\begin{aligned} 0&=\rm{det}(A^2+B^2)\\&=\rm{det}(C\,P^2\,C^{-1}+C\,Q^2\,C^{-1})\\&=\rm{det}(C\,(P^2+Q^2)\,C^{-1})\\&=\rm{det}(P^2+Q^2)\\&=\prod_{i=1}^{n}\left(p_{i}^2+q_{i}^2\right)\,\,(\ast) \end{aligned}$$

    where \(\displaystyle{p_{i}\,,1\leq i\leq n}\) are the diagonial elements of \(\displaystyle{P}\) and \(\displaystyle{q_{i}\,,1\leq i\leq n}\) of \(\displaystyle{Q}\) .

    According to the relation \(\displaystyle{(\ast)}\) we have that : \(\displaystyle{p_{i}^2+q_{i}^2=0}\) for some \(\displaystyle{i\in\left\{1,...,n\right\}}\) .


    But \(\displaystyle{p_{i}\,,q_{i}\in\mathbb{R}}\), so : \(\displaystyle{p_{i}=q_{i}=0}\) and then :

    \(\displaystyle{\rm{det}(A)=\rm{det}(P)=\prod_{i=1}^{n}p_{i}=0=\prod_{i=1}^{n}q_{i}=\rm{det}(Q)=\rm{det}(B)}\) .

    2 votes by Grigorios Kostakos, Apostolos J. Kos

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