# Relation between degree of root of determinant polynomial and rank of the matrix

Let $A=[a_{ij}]$ be an $n \times n$ matrix with $a_{ij}=f_{ij}(x_1,...,x_m)$ where $f_{ij}(x_1,...,x_m)$ is a polynomial in $m$ variables over a finite field $\mathbb{F}_q$.

Let $rank(A)=n$.

Now suppose $(x_i-cx_j)$ divides $determinant(A)$ with $c \in \mathbb{F}_q$ and $rank(A)=n-d$ when $x_i=cx_j$ then does it mean that $(x_i-cx_j)^d$ divides $determinant(A)$ ?

The question can be put more generally where $x_i-cx_j$ can be replaced by an irreducible polynomial $g(x_1,...,x_m)$ where we assume $rank(A)=n-d$ under the assumption $g(x_1,...,x_m)=0$ i.e., we calculate rank of $A$ in quotient space $\mathbb{F}_q[x_1,...x_m]/<g(x_1,...,x_m)>$.

I know that something like above is true for eigenvalues under specific circumstances which is a special case of above. I want to know what happens in general (other than the eigenvalue case).

Thanks

From your more general question I infer that you want to look at the coset of your matrix in the quotient (not at evaluation at specific $x_1,\ldots,x_n\in\mathbb{F}_q$).
Without loss of generality, $c=0$ and $i=n$ (else do a linear change of variables).
Your assumption is that if you consider the coset of this matrix modulo $x_n$, it is a matrix of rank $n-r$ with entries in $\mathbb{F}_q[x_1,\ldots,x_{n-1}]$. If you don't do evaluations, computing ranks can be done over the field $\mathbb{F}_q(x_1,\ldots,x_{n-1})$, where you can make the last $r$ rows of that coset matrix equal to zero by elementary row operations. Lifting that to $\mathbb{F}_q(x_1,\ldots,x_{n-1})[x_n]$, this means that all entries in each of the last rows of the transformed matrix are divisible by $x_n$, so the determinant is divisible by $x_n^r$.
I suppose a similar argument would work for an arbitrary irreducible polynomial where you should localise outside the ideal generated by that polynomial instead of looking at $\mathbb{F}_q(x_1,\ldots,x_{n-1})[x_n]$.