## Equal determinants [entire topic moved to new forum]

Apostolos J. Kos on Wednesday, October 28 2015, 07:32 AM
0

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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• 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 :

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

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