= Singularity forced by an odd-dimensional transpose-inverse identity
{title2=$(CB)^{\mathsf T}+B^{-1}C=0$}
Let $B$ be an <invertible matrix> and $C$ a real square <matrix> of the same odd order. If the displayed identity holds, multiply it on the left by $B$ to obtain $BB^{\mathsf T}C^{\mathsf T}=-C$. Taking <determinants> gives
$$
(\det B)^2\det C=(-1)^d\det C=-\det C.
$$
Thus $((\det B)^2+1)\det C=0$. Since $\det B$ is real, the first factor is positive, forcing $C$ to be a <singular matrix>. Both odd dimension and reality matter: an invertible real <antisymmetric matrix> exists in even dimension, while over the complex numbers a squared determinant can equal $-1$.
Back to article page
