# 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.

**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“.

### More Problems about Determinants

Additional problems about determinants of matrices are gathered on the following page:

Add to solve later

## 1 Response

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