I was working on a proof that the determinant of an antisymmetric ($n\times n$)-matrix is zero if $n$ is odd.
$U_{ij} = (-1)^{i+j}A_{ij}$: algebraic complement to $a_{ij}$, $A_{ij}$ is the subdeterminant $$\det B = \sum_j b_{ij}U_{ij} = \sum_j-b_{ji}U_{ji}=-\sum_j b_{ij}^TU_{ij}^T=-\det B^T$$
But this would mean $\det B = -\det B$ for all $n$ so $\det B$ would be zero for all $n$ which is false.
I know what the correct proof is for $n$ odd but I don't understand what the error in my original proof is. Could someone please point out my error?
$\sum_j b_{ij}U_{ij} = \sum_j-b_{ji}U_{ji}$ is problematic because you know $b_{ij}=-b_{ji}$, but you don't know the relationship between $U_{ij}$ and $U_{ji}$. For example, take $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$; in this example you always have $U_{ij}=-U_{ji}$ so $\sum_j b_{ij}U_{ij} = \sum_jb_{ji}U_{ji}$.