On the linear factors of a polynomial obtained from the determinant of a matrix whose entries are related to Binomial expansion

Question

Consider the polynomial ring $R=\mathbb C[x,y]$.

Consider the matrix $A=\begin{pmatrix} x^5+y^5&5x^5&10x^5&10x^5&5x^5\\5y^5&x^5+y^5 &5x^5&10x^5&10x^5 \\10y^5&5y^5&x^5+y^5&5x^5&10x^5\\10y^5&10y^5&5y^5&x^5+y^5&5x^5\\5y^5&10y^5&10y^5&5y^5&x^5+y^5 \end{pmatrix} \in M_5(R)$.

Also consider the polynomial $p(x,y)=\det (A-I)\in R=\mathbb C[x,y]$.

How to show that $p(x,y) $ can be factored into $5^2=25$ linear polynomials in $x$ and $y$ (over $\mathbb C$) ?

If $x+y=1$, then it can be shown that $1$ is an an eigenvalue of $A$, so the image of $p(x,y)$ in $\mathbb C[x,y]/(x+y-1)$ is $0$; thus $x+y-1$ is a factor of $p(x,y)$; but other than that, I'm unable to say anything else.

I feel I some how have to apply Hilbert Nullstellensatz, but I don't know how.

Answer 1

Define the $\, n\times n\,$ matrix $\, A = \{a_{i, j}\}_{i, j=1}^n \,$ where $\, a_{i, j} = \binom{n}{j-i}x^n + \binom{n}{i-j}y^n. \,$ The matrix $A$ is a special Toeplitz matrix. Let $\, p(x, y) := \det(A-I).\,$ Since $\, x + y - 1 \,$ is a factor, then also $\, z x + w y - 1 \,$ is a factor where $\, z, w \,$ are any pair of $\,n$th roots of unity and there are $\,n^2\,$ pairs.