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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. – alexx May 5 '12 at 21:13
@alex This is not always true, think for example of the matix formed by the vectors $(0,1)$ and $(1,0)$ in column form. – William May 5 '12 at 21:17
This has been crossposted to MO‌​. Perhaps someone with 50+ MO rep (I'm 1 point short!) could inform them? – Alex Becker May 11 '12 at 17:56

Take $A+D=\begin{pmatrix} 1& a& b\\ a& 1& c\\ b& c& 1 \end{pmatrix}$ with $a$, $b$, $c$ real.

Now, the entries of $D$ are 1 and the determinant of $A+D$ is $1+2abc-a^2-b^2-c^2$.

You want to recover $\det A=2abc$.

If you regard $a=b=1$ and $c=0$, then $\det A+D=-1$ and $\det A=0$.

But if you regard $a=b=1$ and $c=2$, then you also get $\det A+D=-1$, but $\det A=4$.

This example shows that your information is not sufficient to distinguish the two cases. So, in general it is impossible to recover the determinant of the original matrix $A$.

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