Home » Uncategorized » Equal determinants

Equal determinants

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

Solution

The key point here is that the matrices are simultaneously diagonisable. Thus , there exists an invertible matrix C\in\mathbb{M}_{n}(\mathbb{R}) and diagonisable matrices P\,,Q\in\mathbb{M}_{n}(\mathbb{R}) such that

A=C\,P\,C^{-1}\,\,,B=C\,Q\,C^{-1}

Hence

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

where p_{i}\,,1\leq i\leq n are the diagonial elements of P and q_{i}\,,1\leq i\leq n of Q. According to (1) we have that

p_{i}^2+q_{i}^2=0 \quad \text{for some} \; i\in\left\{1,...,n\right\}

But p_{i}\,,q_{i}\in\mathbb{R}, so p_{i}=q_{i}=0 and finally we conclude that:

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

Read more

Leave a comment

Donate to Tolaso Network