# Determinant of a real skew-symmetric matrix is square of an integer

Let $A$ be a real skew-symmetric matrix with integer entries. Show that $\operatorname{det}{A}$ is square of an integer.

Here is my idea: If $A$ is skew-symmetric matrix of odd order, then $\operatorname{det}{A}$ is zero. So, take $A$ to be of even order and non-singular. Since all the eigenvalues of $A$ are of the form $ia$ and its conjugate (where $a$ is real number), we see that $\operatorname{det}{A}$ is square of a real number. But I am not getting how to show it is square of an integer.

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en.wikipedia.org/wiki/Pfaffian – m_t_ Jul 2 '12 at 18:03
I deleted my answer since it doesn't make sense. Thanks to @JasonDeVito. – user2468 Jul 2 '12 at 18:29
Sorry,I need more explanation.I did not get the idea. – Shraddha Srivastava Jul 2 '12 at 18:55
The Pfaffian is a polynomial function of the matrix entries – PAD Jul 2 '12 at 19:45

A proof by induction is given in David J. Buontempo, The determinant of a skew-symmetric matrix, The Mathematical Gazette, Vol. 66, No. 435, Mar., 1982, Note 66.15, pages 67-69. If you have access to jstor, it's here. The proof does not depend on the Pfaffian.

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For a skew symmetric $A$, $\det(A)={\rm pfaffian}(A)^2$ where pfaffian is an integral polynomial function of the entries of the matrix $A$. For the case of an integer matrix the pfaffian is therefore an integer. Hence the result you want.

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Yes, and that information is already available at the Wikipedia/Pfaffian link in the first comment on the question. – Gerry Myerson Jul 6 '12 at 0:42