One can try to generalize the Vandermonde determinant in the following direction: Let $A$ be any symmetric $n$-order square matrix. Consider its powers' diagonal elements $(A^k)_{ii}$ and construct square matrix $V(A)$ who's rows are $((A^k)_{11}, (A^k)_{22},...,(A^k)_{nn})$, where $k=0,1,2,...,n-1$ and it is assumed that $((A^0)_{11}, (A^0)_{22},...,(A^0)_{nn})=(1,1,1,...,1)$, that is $A^0$ is the identity matrix. The determinant $\det(V(A))$ can be called the Vandermonde determinant of $A$. My question is: Has this determinant been considered somewhere regarding non-singularity of $V(A)$? May be it is considered at least in symmetric tridiagonal $A$ case? I am interested in a very particular case of symmetric tridiagonal $A$: Its diagonal elements are $1,2,2,...,2$, upper(lower) adjacent diagonal consists only of $-1$. I have a conjecture that its Vandermonde determinant is $\pm 1$, depending on $n$. But I have no proof of it.
Yes, the conjecture is true. It can be proved by induction on $n>5$. The main idea is that somewhere in the middle of $V(A)$ there are two adjacent columns with difference $(0,0,...,0,1)$.
Nevertheless investigation of $V(A)$ even for symmetric tridiagonal matrices is interesting. In particular: Does $det(A)=0$ imply $det(V(A))=0$?
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