The Determinant of a Skew-Symmetric Matrix is Zero

Problem 473
Prove that the determinant of an $n\times n$ skew-symmetric matrix is zero if $n$ is odd.
Contents
Definition (Skew-Symmetric)
A matrix $A$ is called skew-symmetric if $A^{\trans}=-A$.
Here $A^{\trans}$ is the transpose of $A$.
Proof.
Properties of Determinants
We will use the following two properties of determinants of matrices.
For any $n\times n$ matrix $A$ and a scalar $c$, we have
- $\det(A)=\det(A^{\trans})$,
- $\det(cA)=c^n\det(A)$.
Main Part of the Proof
Suppose that $n$ is an odd integer and let $A$ be an $n \times n$ skew-symmetric matrix.
Thus, we have
\[A^{\trans}=-A\]
by definition of skew-symmetric.
Then we have
\begin{align*}
\det(A)&=\det(A^{\trans}) && \text{by property 1}\\
&=\det(-A) && \text{since $A$ is skew-symmetric}\\
&=(-1)^n\det(A) && \text{by property 2}\\
&=-\det(A) && \text{since $n$ is odd}.
\end{align*}
Therefore, it yields that $2\det(A)=0$, and hence $\det(A)=0$.
Comment.
The result implies that every odd degree skew-symmetric matrix is not invertible, or equivalently singular.
Also, this means that each odd degree skew-symmetric matrix has the eigenvalue $0$.
Related Question.
The eigenvalues of a real skew-symmetric matrices are of the special form as in the next problem.
(b) Prove that the rank of $A$ is even.
For a proof, see the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even“.
More Problems about Determinants
Additional problems about determinants of matrices are gathered on the following page:

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