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

I know that in general, a skew-symmetric matrix with indeterminate elements has a determinant that can be written as a square of some multivariable polynomial. How to prove this?

And if I do not know anything about the Pfaffian, can I prove the statement that a skew-symmetric matrix with integer entries has a determinant that is a square of some integer? I mean if someone knows how to prove it without using the Pfaffian?


marked as duplicate by user7530, Hagen von Eitzen, Chris Eagle, Davide Giraudo, sdcvvc Jan 13 '13 at 11:29

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This is for the second part, a skew-symmetric matrix with integer entries

First, if $n$ is odd, then since $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$, so $\det(A)= 0$, which is the square of an integer.

Now for $n$ even, we proceed by induction, and will show the statement is true over the rationals. Base case $n=2$ is obvious.

Inline Edit: Induction step. Suppose we want to show it for some $n=2k+2$. Consider the entries. We already know that the diagonals satisfy $a_{i, i} = 0$. If all the other entries are also 0, then we are done. Otherwise, WLOG we have $a_{1,2}\neq 0$. We will proceed to calculate the determinant of $A$. end edit

Using only row operations, where we add a rational linear multiple of the second row to the others, we can make the first column to be $(0, a_{2, 1}, 0, 0, \ldots, 0)^T$. Specifically, for row $k\neq 1, 2$, we have $b_{k, i} = a_{k, i} - \dfrac {a_{k, 1}}{a_{2, 1}} a_{2,i}$. Now, we use column operations, where we add a rational linear multiple of the second column to the others, and we can make the first row to be $(0, a_{1, 2}, 0, 0, \ldots, 0)$. Specifically, for column $k\neq 1, 2$, we have $c_{i, k} = b_{i, k} - \dfrac {b_{1, k}}{b_{1, 2}} b_{i, 2}$. Then, $\det(A) = a_{1, 2}^2 \det(C)$, and so it remains to show that $C$ is still a skew symmetric matrix.

For rows and columns 1 and 2, nothing changed throughout, (so those entries are skew symmetric).
For $i=k$, $b_{i,i} = 0- \dfrac {a_{i,1}}{a_{2,1}}a_{2,i}$, $c_{i,i} = b_{i,i} - \dfrac {b_{1,i}}{b_{1, 2}} b_{i, 2} = (0 - \dfrac {a_{i,1}}{a_{2,1}}a_{2,i})- \dfrac {a_{1,i}}{a_{2,i}} a_{i,2}=0$

For $i\neq k$, $c_{i,k} = b_{i, k} - \dfrac {b_{1, k}}{b_{1, 2}} b_{i, 2} = b_{i, k} - \dfrac {a_{1, k}}{a_{1, 2}} a_{i, 2}= (a_{i, k} - \dfrac {a_{i, 1}}{a_{2, 1}} a_{2,k})- \dfrac {a_{1, k}}{a_{1, 2}} a_{i, 2}$. Hence, it follows that $c_{i,k}=-c_{k,i}$.

Hence, $C$ is a skew symmetric matrix.

Now, $A$ has an integer determinant, which is also the square of a rational number, hence is a perfect square.

  • $\begingroup$ @adamW I came up with this looking at the 4 by 4 case. I didn't feel that identifying 0's would help, since it doesn't really affect the general identity. In fact, there was no issue go dividing by 0, since we only divided by $a_{1, 2}$ and $a_{2, 1}$ which we can choose to be non-zero. Furthermore, with the definition of $B$, the 0 entries on the diagonal will have disappeared. $\endgroup$ – Calvin Lin Jan 13 '13 at 18:37
  • $\begingroup$ You may have misunderstood, I only meant that I had to realize the zeros were there to understand your statement "expanding along that row will give us the result via the induction hypothesis" $\endgroup$ – adam W Jan 13 '13 at 18:53
  • $\begingroup$ @adamW Ah, I see. Hm, but I already stated that $a_{i,i}=0$. Will add a line for clarity. $\endgroup$ – Calvin Lin Jan 13 '13 at 18:54
  • $\begingroup$ @adamW The statement was wrongly written. It should be, if there is $a_{I, J} \neq 0$, then we proceed as below. Notice that in the subsequent paragraphs, we only divided out by $a_{1,2}$. Otherwise, if all $a_{I, J} = 0$ the determinant is 0. $\endgroup$ – Calvin Lin Jan 13 '13 at 19:28
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    $\begingroup$ +1 very nice. Just a comment: the entries of $C$ will not be integers but this is not a problem. If $r$ is a common denominator then $r^n$ is a square and $r^nC$ has integer entries. $\endgroup$ – P.. Jan 13 '13 at 20:25

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