If $n$ is odd and $a_{ij} = -a_{ji}$ for all $i,j\in\{1,...,n\}$, then $A$ is not invertible. Does the statement hold also when $n$ is even?

Question

I have no clue how to do that task. We currently are doing Eigenvalue, Eigenvector and diagonaziable matrices. Do you guys have any ideas how to do this task? I am very sorry for the formating, I cannot get the hang of it.

Translation of the question: $K$ is a field, $n$ a natural number and $A = (a_{ij}), i,j = 1,...,n \in Mat(n \times n, K)$. Prove this statement: If $n$ is odd and $a_{ij} = -a_{ji}$ for all $i,j \in \{1,...,n\}$, then $A$ is not invertible. Does the statement also apply to the case when $n$ is even?

Thanks in Advance!

Answer 1

Consider $$ \begin{pmatrix}0&4\\-4&0\end{pmatrix}$$ which is invertible.

Answer 2

As Jyrki Lahtonen has noted, the simplest counterexample to an even-$n$ attempt to generalize the theorem is $\left(\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array}\right)$. The reason the theorem only works for odd $n$ is because $A^T=-A$ implies $\det A=\det A^T=(-1)^n \det A$, which implies $\det A=0$ provided $(-1)^n=-1$.