Is every odd order skew-symmetric matrix singular? 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?
Yes, that holds, since: $$\det A=\det{(-A^T)}=(-1)^{odd}\det{A^T}=-\det A,$$ from where we get $\det{A}=0$.
Consider the example
$$
\begin{bmatrix}
    0 &amp; i &amp; -3\\
    -i &amp; 0 &amp; 2i\\
    3 &amp; -2i &amp; 0\\
\end{bmatrix}
$$