10
$\begingroup$

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.

$\endgroup$
  • 7
    $\begingroup$ en.wikipedia.org/wiki/Pfaffian $\endgroup$ – Matthew Towers Jul 2 '12 at 18:03
  • $\begingroup$ I deleted my answer since it doesn't make sense. Thanks to @JasonDeVito. $\endgroup$ – user2468 Jul 2 '12 at 18:29
  • $\begingroup$ Sorry,I need more explanation.I did not get the idea. $\endgroup$ – user51266 Jul 2 '12 at 18:55
  • $\begingroup$ The Pfaffian is a polynomial function of the matrix entries $\endgroup$ – PAD Jul 2 '12 at 19:45
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is it possible to include the images of the 3 pages here? $\endgroup$ – StubbornAtom Mar 18 '17 at 12:30
  • $\begingroup$ I'm certainly not going to do that. If you have access to a library, I'm sure they can get it for you on Interlibrary Loan. $\endgroup$ – Gerry Myerson Mar 19 '17 at 12:13
  • $\begingroup$ Never mind, I got it legally. $\endgroup$ – StubbornAtom Mar 19 '17 at 13:51
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy