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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.

Thankyou very much.

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    Well, if you assume that $A$ is a Hermitian matrix whose diagonal entries are all zeros, then $A$ has an eigenvalue of $0$, and consequently its determinant is $0$. So I've just ``recovered'' the determinant of $A$ for you and the determinants of $A+D$ and the entries of $D$ were not necessary.2012-05-05
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    @alex This is not always true, think for example of the matix formed by the vectors $(0,1)$ and $(1,0)$ in column form.2012-05-05
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    This has been crossposted to [MO](http://mathoverflow.net/questions/96691/determinant-of-the-sum-of-matrices). Perhaps someone with 50+ MO rep (I'm 1 point short!) could inform them?2012-05-11

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