Let us denote the determinant of a matrix $A=((a_{ij}))_{i,j=1}^n$ as $\displaystyle \det_{i,j=1}^n a_{ij}$. Let $\delta_{ij}$ be the kronecker delta function and suppose $a_{ij}=a_{ji}$ for all pair $(i,j)$. Then is the following true? $$\det_{i,j=1}^n (\delta_{ij}-a_{ij})=1+\sum_{k=1}^n(-1)^k\sum_{1\le v_1<v_2<\cdots<v_k\le n} \det_{i,j=1}^k a_{v_iv_j}$$
If it is some standard determinant expansion, I am not aware of it. Any help/suggestions.
The one that you write is just the standard Laplace expansion of the determinant on the left hand side, written as a inhomogeneous polynomial in the $a_{ij}$.
Indeed, $det( I - A)$ is an inhomogeneous polynomial in the $a_{ij}$.
I give an idea of what happens in the expansion:
-the term of degree $0$ is obtained from the term of degree $0$ in the product of the diagonal elements of $I-A$, therefore it is $1$;
-the term of degree $1$ is obtained by taking the $1$ from $n-1$ diagonal entries and multiply them by the variable $a_{ii}$ in the remaining diagonal entry, in all the possible ways (that are $\binom{n}{1} = n$); this gives rise to the summand with $k=1$;
-the term of degree $2$ is obtained by taking $n-2$ diagonal $1$'s multiplied by the $2 \times 2$ minor obtained from the remaining two rows/columns; you have $\binom{n}{2}$ possible ways to do that, as many as the $2 \times 2$ minors centered on the diagonal; this gives rise to the summand with $k=2$;
higher terms are more complicated but the idea is the same; the term of degree $k$ is obtained by taking $n-k$ diagonal $1$'s multiplied by the complementary $k\times k$ minor centered on the diagonal.