Linear Algebra

The Determinant of a Skew-Symmetric Matrix is Zero

by Yu · Published 06/23/2017 · Updated 12/02/2017

Problem 473

Prove that the determinant of an $n\times n$ skew-symmetric matrix is zero if $n$ is odd.

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Contents

Problem 473
Definition (Skew-Symmetric)
Proof.
Related Question.
- More Problems about Determinants

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.

Problem. (a) Prove that each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number.
(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“.

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