What can be a explicit formula for the determinant of the following matrix? $$A={\begin{pmatrix} 0 & 1 & & & & & &1\\ 1 & 0 & 1 & & & & &\\ & 1 & 0 & 1 & & & &\\ & & & \ddots & & & &\\ & & & 1 & 0 & 1 & & \\ & & & & 1 & 0 &1&\\ & & & & & 1 &0&1\\ 1 & & & & & &1&0 \end{pmatrix}}\in \textrm{M}(n,\mathbb{R}).$$ The white spaces outside the tridiagonal are $0$.
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Similar: https://math.stackexchange.com/questions/2477418/finding-eigenvalues-in-almost-tridiagonal-matrix – Hans Lundmark Apr 16 '21 at 08:42
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https://en.wikipedia.org/wiki/Circulant_matrix – Martin R Apr 16 '21 at 08:47
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Also similar: https://math.stackexchange.com/q/81016/42969 – Martin R Apr 16 '21 at 08:52
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Let $B$ be the matrix corresponding to the endomorphism $e_i \longmapsto e_{i+1 \pmod n}$, so that $A=B+B^{-1}$. The complex eigenvalues of $B$ are simple, they’re the elements of the set $\mu_n$ of $n$-th roots of unity. Thus $\det{A}=\prod_{\omega \in \mu_n}{\omega+\omega^{-1}}=e^{2i\pi/n \cdot n(n+1)/2} \prod_{k=0}^{n-1}{(e^{4ik\pi/n}+1)}$.
Now, if $n$ is odd, the RHS is $(-1)^n\prod_{k=0}^{n-1}{((-1)-e^{2ik\pi/n})}=(-1)^n((-1)^n-1)=-(-2)=2$.
If $n=2m$, the RHS is $-\prod_{k=0}^{n-1}{(1+e^{4i\pi/n})}=-\left(\prod_{k=0}^{m-1}{(1+e^{2i\pi/m})}\right)^2=-((-1)^m((-1)^m-1))^2$ so the determinant is equal to zero if $n$ is divisible by $4$ and $-2$ if $n$ is even and not divisible by $4$.
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