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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.

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This is likely more appropriate for – Samuel Reid May 11 '12 at 21:28
This had been posted on 6 days prior:… – Jonas Meyer Jan 4 '13 at 6:39

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.

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