Let $A$ be a $n\times n \ (n\geq{0})$ matrix such that $A^T=-A$. For what values of $n$ is $\det(A)=0$? Using one of the properties of the determinant, $\det(A)=0$ for $n=2$.
If $n=2$ then the determinant of $A$ will be zero because `The determinant of a matrix with two identical rows is 0.
Am I right in reasoning this way, or am I missing something?
 A: HINT
Start with noticing that
\begin{align*}
A^{t} = -A & \Rightarrow \det(A^{t}) = \det(-A)\\\\
& \Rightarrow \det(A) = (-1)^{n}\det(A)\\\\
& \Rightarrow \det(A)(1 - (-1)^{n}) = 0
\end{align*}
Can you take it from here?
A: I'm assuming that the characteristics of the field is not 2.
Because the determinant is a multilinear function in the rows, than for any scalar $\lambda$ the determinant of an $n\times n$ matrix uphold $\det (\lambda\cdot A)=\lambda^n\cdot\det(A)$. Thus we can conclude that if $A^T=-A$ then $\det(A)=(-1)^n\cdot\det(A)$. If $n$ is odd it follows that $\det(A)=-\det(A)$ so it has to be that $\det(A)=0$ (because the characteristics of the field is not 2). If $n$ is even then the matrix of the form
\begin{pmatrix}
0 & 0 & \cdots & 0 & -1\\
0 &  &  & -1 & 0\\
\vdots & &  &  & \vdots\\
0 & 1 &  &  & 0\\
1 & 0 & \cdots & 0 & 0
\end{pmatrix}
(-1 on the upper half of the secondary diagonal and 1 on the lower half of the secondary diagonal) will have determinant that is not zero because if we will expand the determinant with its minors on the secondary diagonal it will be a product of non zero numbers (in the case where $n$ is odd this matrix wont be possible because it will had to have a zero in the middle of the secondary diagonal).
