Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
6  
This is likely more appropriate for math.stackexchange.com –  Samuel Reid May 11 '12 at 21:28
    
This had been posted on math.stackexchange.com 6 days prior: math.stackexchange.com/questions/141499/… –  Jonas Meyer Jan 4 '13 at 6:39

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.