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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. The question is that whether every odd order skew-symmetric with complex entries is singular?

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

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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\ $

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  • $\begingroup$ But the problem here is that the entries are complex. Is it true for complex entries as well? $\endgroup$ – Shahnawaz Ahmad Nov 18 '16 at 12:05
  • $\begingroup$ Yes, it is also true for complex entries. The proof does not assum real entries. $\endgroup$ – Peter Nov 18 '16 at 12:47
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Consider the example $$ \begin{bmatrix} 0 & i & -3\\ -i & 0 & 2i\\ 3 & -2i & 0\\ \end{bmatrix} $$

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  • $\begingroup$ I mean a 3 by 3 matrix with rows (0 i -3), ( -i 0 2i) and (3 -2i 0). $\endgroup$ – Shahnawaz Ahmad Nov 18 '16 at 12:21

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