Proof the following properties of antisymmetric matrices:

Question

a) If A is antisymmetric of order $p$, then $\det(A)=(-1)^pdet(A)$

b) If A is odd order antisymmetric, then $\det(A)=0$

Remembering that if A is antisymmetric, $A ^ T = -A$

How could I prove the two propositions?

Answer 1

Assuming $p$ is the matrix dimension:

(a) It's well-known that $\det(A) = \det(A^{T})$ and that $\det(kA)=k^p\det(A)$ for any scalar $k$. For any antisymmetric (skew-symmetric) matrix we have $A=-A^{T}$. Therefore

$$\det(A) = \det(A^{T}) = \det(-A)=(-1)^p\det(A)$$

(b) This is just a corollary of (a). Since $p$ is odd, we have $(-1)^p = -1$ and thus

$$\det(A) =(-1)^p\det(A) = -\det(A)$$

and it follows that $\det(A) = 0$, i.e. the matrix is singular.