Is the number of invertible 4×4 circulant matrices over the ring of integers Z finite? I am looking for a reference which discusses this case.
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1$\begingroup$ It seems that there are four infinite families: those with first row $(n\pm1,k,-n,-k)$ and those with first row $(n,k\pm1,-n,-k)$, with $n,k\in\Bbb{Z}$. $\endgroup$– Olivier BégassatCommented Aug 17, 2022 at 11:58
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$\begingroup$ I tried it for k=2 and n=3 it does not work..the determinant is not 1 or _1 so it is not invertible in Z?Am I wrong? $\endgroup$– KatyCommented Aug 17, 2022 at 12:05
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1$\begingroup$ As far as I can tell this should work because the complex eigenvalues of a 4x4 circulant matrix with first row $(a,b,c,d)$ are $(a+c) + (b+d)$, $(a+c) - (b+d)$, $(a+c) +i(b+d)$ and $(a+c) - i(b+d)$ which will all be in $\{\pm1, \pm i\}$ unless I'm mistaken. $\endgroup$– Olivier BégassatCommented Aug 17, 2022 at 12:16
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1$\begingroup$ My bad, I forgot some minus signs... The eigenvalues are $(a+c) + (b+d)$, $(a+c) - (b+d)$, $(a-c) -i(b-d)$ and $(a-c) + i(b-d)$ so you'll need ($c = a \pm 1$ and $b=d$) or ($c = a$ and $d=b\pm 1$). In the first case you also need $\pm 1 + 2a + 2b = \pm 1$ and $\pm 1 + 2a - 2b = \pm 1$ so that it seems only a few possiibilities for a and b are allowed. Something analoguous holds in the second case. There are thus only finitely many possibilities. $\endgroup$– Olivier BégassatCommented Aug 17, 2022 at 12:35
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$\begingroup$ Thanks for the explanation $\endgroup$– KatyCommented Aug 17, 2022 at 12:40
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1 Answer
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There are only finitely many such matrices. Let us consder the circular matrix say $A$, with the first row $(i,j,k,m)$. Then $det(A)=FGH$, where $$ F=i-j+k-m,\; G=i+j+k+m,\;H=(i-k)^2+(j-m)^2 $$ We need to have $det(A)=\pm 1$. Thus, we need to play with the following system of equations \begin{align*} (A) \quad &F=G=H=1,\\ (B) \quad &F=G=-1, H=1,\\ (C) \quad &F=-1, G=1, H=1,\\ (D) \quad &F=1, G=-1, H=1. \end{align*} Is is a simple excercise to prove that each of the above systems has only finitely many solotions (under the condition that $i, j, k, m\in\mathbb{Z}$).
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1$\begingroup$ I believe that one of $i,j,k,m$ must be $\pm 1$ and all the rest 0. So 8 solutions in total. $\endgroup$ Commented Aug 18, 2022 at 9:37
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