Let $a_1,...,a_{2n} \in \mathbb C$ and $A=[b_{ij}]\in M(2n,\mathbb C)$ such that $A^T=-A$ and $b_{ij}=a_ia_j,\forall i<j$.
Can we find a nice expression for determinant of $A$?
Let $a_1,...,a_{2n} \in \mathbb C$ and $A=[b_{ij}]\in M(2n,\mathbb C)$ such that $A^T=-A$ and $b_{ij}=a_ia_j,\forall i<j$.
Can we find a nice expression for determinant of $A$?
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By definition, $A=DKD$ where $D=\operatorname{diag}(a_1,\ldots,a_{2n})$ and $K$ is the skew symmetric matrix whose strictly upper triangular part is one. It is easy to prove that $\det(K)=1$. Hence $\det(A)=\prod_{i=1}^{2n}a_i^2$.