Let $A$ be an $n\times n$ matrix and $A^t$ be its transpose.
I know how to show $\mathrm{det}(A-A^t)=(-1)^n\mathrm{det}(A^t-A)$.
I would like to know how to show $\mathrm{det}(A-A^t)=(-1)^{\mathrm{rank} A}\mathrm{det}(A^t-A)$.
The strategy I took is as follows. If I can find matrices $U$ and $V$ such that $UAV$ and $UA^tV$ are in Row echelon form, then I can prove it but I also don't know how to construct these.
Any help is appreciated. Thank you in advance.
Best Answer
What you want to prove is not true. Take
$$A = \left(\begin{array}{cc} 1 & 3 \\ 2 & 6 \end{array}\right)$$
for which $\operatorname{rank}A=1$ and $\det(A-A^t)=\det(A^t-A)=1$,
because
$$A^t = \left(\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}\right)$$ $$A^t-A = \left(\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right)$$ .