Jacobi's Theorem Any skew-symmetric matrix of odd order has determinant equal to $0$ .

Proof. Suppose $A$ is an $n\times n$ square matrix. For the determinant, we then have $\det A = \det A^T$ , and $\det (-A) = (-1)^n \det A$ . Thus, since $n$ is odd, and $A^T=-A$ , we have $\det A = -\det A$ , and the theorem follows.

Remarks

1. According to [1], this theorem was given by Carl Gustav Jacob Jacobi (1804-1851) [2] in 1827.
2. The $2\times 2$ matrix shows that Jacobi's theorem does not hold for $2\times 2$ matrices. The determinant of the $2n\times 2n$ block matrix with these $2\times 2$ matrices on the diagonal equals $(-1)^n$ . Thus Jacobi's theorem does not hold for matrices of even order.
3. For $n=3$ , any antisymmetric matrix $A$ can be written as $$ A = \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{pmatrix} $$ for some real $v_1,v_2,v_3$ , which can be written as a vector $v=(v_1,v_2,v_3)$ . Then $A$ is the matrix representing the mapping $u\mapsto v\times u$ , that is, the cross product with respect to $v$ . Since $Av=v\times v=0$ , we have $\det A=0$ .

Bibliography

1

H. Eves, Elementary Matrix Theory, Dover publications, 1980.

2

The MacTutor History of Mathematics archive, Carl Gustav Jacob Jacobi