# 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 $\det{A}=0$.

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\$

• But the problem here is that the entries are complex. Is it true for complex entries as well? Nov 18 '16 at 12:05
• Yes, it is also true for complex entries. The proof does not assum real entries. Nov 18 '16 at 12:47

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

• I mean a 3 by 3 matrix with rows (0 i -3), ( -i 0 2i) and (3 -2i 0). Nov 18 '16 at 12:21
• If this was intended to be a counterexample, then it is not! The determinant of this matrix is zero. May 6 at 15:03