I'm working on a proof right now, and the question asks about an invertible skew-symmetric matrix. How is that possible? Isn't the diagonal of a skew-symmetric matrix always $0$, making the determinant $0$ and therefore the matrix is not invertible?

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    $\begingroup$ Having vanishing diagonal entries means the trace is always zero, but the determinant need not necessarily be zero. Consider $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$. $\endgroup$ – Dustan Levenstein May 3 '15 at 0:37
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    $\begingroup$ To jump a bit forward: odd-order skew-symmetric matrices are necessarily singular, but even-order ones don't have to be. $\endgroup$ – J. M. is a poor mathematician May 3 '15 at 0:46

The diagonal of a skew-symmetric matrix is always $0$ does not mean that its determinant be $0$. Look at following example:

$det\left[ \begin{array}{} 0 & 1 \\ -1 & 0 \\ \end{array} \right]=1 $

Its inverse is: $\left[ \begin{array}{} 0 & -1 \\ 1 & 0 \\ \end{array} \right] $


No, the diagonal being zero does not mean the matrix must be non-invertible. Consider $\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$. This matrix is skew-symmetric with determinant $1$.
Edit: as a brilliant comment pointed out, it is the case that if the matrix is of odd order, then skew-symmetric will imply singular. This is because if $A$ is an $n \times n$ skew-symmetric we have $\det(A)=\det(A^T)=det(-A)=(-1)^n\det(A)$. Hence in the instance when $n$ is odd, $\det(A)=-\det(A)$; over $\mathbb{R}$ this implies $\det(A)=0$.


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