Is every odd order skew-symmetric matrix singular?

Question

We call a square matrix $A$ a skew-symmetric matrix if $A=-A^T$. A matrix is said to be singular if its determinant is zero. Is every odd order skew-symmetric matrix with complex entries singular?

Answer 1

This is actually the case :

Suppose, $A$ is an $n\times n$-matrix.

We have $$\det(A)=\det(-A^T)=(-1)^n\cdot \det(A^T)=(-1)^n\cdot \det(A)$$

Since $n$ is odd, we can conclude $\ \det(A)=-\det(A)\ $ implying $\ \det(A)=0\ $

Answer 2

Yes, that holds, since: $$\det A=\det{(-A^T)}=(-1)^{odd}\det{A^T}=-\det A,$$ from where we get $\det{A}=0$.

Answer 3

Consider the example $$ \begin{bmatrix} 0 & i & -3\\ -i & 0 & 2i\\ 3 & -2i & 0\\ \end{bmatrix} $$