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I got problem in determining the determinant of specific circulant matrix $C$ formed by shifting the vector $1\cdots101\cdots10\cdots0$. The number of $1$'s in the first sequence of $1$'s is $k$ and the one of the second sequence of $1$'s is $k+1$. These two sequences of $1$'s are separated by exactly one $0$. The number of $0$'s at the end is arbitrary. My conjecture is that the determinant of $C$ is always odd. I do wish some one would like to help me. Thank you.

the_candyman
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I Nengah Suparta
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1 Answers1

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This is only an answer in the case where there are no zeros at the end (i.e., one zero in each row and column). Perhaps you can extend the method.

In this case, let $C=(c_{i,j})$ be the matrix in question. Of course $C$ is $n\times n$ where $n=2k+2$. We want to show that $\det(C)\equiv 1 \pmod 2$.

By definition, $$\det(C) \equiv \sum_{\sigma\in S_n} c_{1,\sigma(1)}c_{2,\sigma(2)}\cdots c_{n,\sigma(n)} \pmod 2.$$

Note each summand is $0$ or $1$ of course, and the summand corresponding to permutation $\sigma$ is $0$ if and only if at least one of the $c_{i,\sigma(i)}=0$. Let $\Omega$ be the set of such $\sigma$. We'll use inclusion-exclusion to count $|\Omega|$.

Let $D$ be the coordinates of the entries of $C$ which are zero (so $|D|=n$). Of course, for any $k$ coordinates $(i_1,j_1),\ldots,(i_k,j_k)\in D$, there are $(n-k)!$ permutations $\sigma\in S_n$ with $\sigma(i_\ell)=j_\ell$ for all $\ell=1,\ldots,k$. It follows that \begin{eqnarray*} |\Omega| &=& n(n-1)! - {n\choose 2}(n-2)!+{n\choose 3}(n-3)!-\cdots + (-1)^n {n \choose n}(n-n)!\\&=&n!-\frac{n!}{2!}+\frac{n!}{3!}-\cdots + (-1)^n.\end{eqnarray*}

But $n=2k+2$ is even, so for $1\leq p <n$, $\frac{n!}{p!}$ is even, and thus $|\Omega|$ is odd. Therefore, $n!-|\Omega|$ is odd and we see that in $\det(C)$ there are an odd number of $1$'s, and so $\det(C)\equiv 1 \pmod 2$ as desired.

Casteels
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  • Doh, should have looked at the "related" sidebar. This http://math.stackexchange.com/questions/81016/determinant-of-a-specific-circulant-matrix-a-n?rq=1 is basically the same as the special case I answered, up to sign, with way better answers! – Casteels Nov 28 '13 at 10:41
  • this special case is absolutely the same as math.stackexchange.com/questions/81016 as you answered. Even you gave a very simple answer on it. But as you suggested me, probably on the current answer will appear things which we can generalize. My further question is, is there any dependence of $|\Omega|$ to $|D|$? I could not see that. – I Nengah Suparta Nov 28 '13 at 14:23
  • I'm not sure. Honestly to generalize my answer might be too horrible. Perhaps a better thought is that since the determinant and permanent are the same modulo 2, you could try interpreting this question graph theoretically. – Casteels Nov 28 '13 at 14:42
  • Although the matrices are absolutly regular (in therms of the number of $0$'s in each row and each coulumn), but for me it is still very hard to find the number (in fact $the$ $parity$) of permutations $\sigma\in S_{n}$ such that $\Sigma _{\sigma\in S_n}=0$. It should be $odd$ in my opinion. Do you have any idea? – I Nengah Suparta Nov 28 '13 at 18:13
  • No not really, just as I said you can think about the permanent as counting the number of perfect matchings in a $2k+1$-regular bipartite graph. This is a hard problem in general, but perhaps your graph has some special structure. I haven't actually looked at it. – Casteels Nov 28 '13 at 18:17