Let $D$ be a diagonal matrix and $A$ a Hermitian one. Is there a nontrivial way to calculate the determinant of $A$ from the determinant of $A+D$ and the entries of $D$?
It can be assumed that the diagonal entries of $A$ are all zeros.
Thank you very much.
Let $B=A+D$. With $B_1,\dots,B_n$ the columns of $B$, $d_1,\dots,d_n$ the diagonal $D$ $$ \det A=(B_1-d_1e_1)\wedge\dots\wedge (B_n-d_ne_n) $$ so that $\det A$ is an explicit polynomial in $d$, whose constant coefficient is $\det B$ and the term of highest degree is $(-1)^nd_1\dots d_n$. For instance, the coefficient of $d_1$ is $ -e_1\wedge B_2\wedge \dots\wedge B_n, $ and all the coefficients can be expressed explicitly.
