Inversible Antisymmetric Matrix

Question

Let $A$ be a $n \times n$ antisymmetric matrix. Show that if $A$ is invertible then $n$ is even.

I guess that this means that $\det(A)$ is zero when $n$ is odd and non-zero when it is even but I dont know how to show this.

Thanks

Answer 1

We have

det$(A) = $det$(-A^T) = (-1)^n$det$(A^T) = (-1)^n$det$(A)\Rightarrow (-1)^n =1 \Rightarrow n$ is even.

Answer 2

An antisymmetric matrix $A$ is a matrix with

$$A^T = -A.$$

Taking the determinant on both sides yields

$$\det (A^T) = \det (-A).$$

How could you simplify the RHS?

Answer 3

well, we know that $$|A|=|A^t|=|-A|=(-1)^n|A|$$ and so if $n$ is odd we get that $|A|=0$.