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I needed to find a lower bound on $|\det(A+B)|$ where $|.|$ is the absolute value operator.

Because I was unable to get such a bound so I was trying to guess a bound and prove it.

But $||\det(A)|-|\det(B)||$ does not seem to work.

If $\sigma_1$, $\sigma_2$, ..., $\sigma_n$ are singular values of $A$, and $\mu_1$, $\mu_2$, ..., $\mu_n$ are singular values of $B$, then $|(\sigma_1-\mu_1)(\sigma_2-\mu_2)\cdots(\sigma_n-\mu_n)|$ does not seem to work as the lower bound either.

By "does not seem to work" I mean it is not the lower bound of $|\det(A+B)|$.

Any pointers on how to proceed will be helpful. A bound in case there is a significant difference between the singular values of $A$ and $B$ may also work.

The only case that I can solve for is when both $A$ and $B$ are positive semi definite, where $|\det(A+B)|=\det(A+B)\geq \det(A)+\det(B)=|\det(A)|+|\det(B)|.$

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The trivial solution would be: the lower bound of $|\det(A+B)|$ is $0$. Here is a bit more general solution: under some conditions $\det(A+B) \geq \det(A) + \det(B)$ (see https://mathoverflow.net/questions/65424/determinant-of-sum-of-positive-definite-matrices). If $\det(A+B) \geq 0$, this means $|\det(A+B)| = \det(A+B) \geq \det(A) + \det(B)$. This solution is more informative then the trivial one (as longs as $\det(A) + \det(B) >0$). The Wikipedia page on the Matrix determinant lemma might help aswell.

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  • $\begingroup$ Thanks, the inequality you say only works for positive semi definite matrices. I think I was aware of that when I posted the question 11 months ago. The question is intended for the general case. $\endgroup$ – jinkx Jun 3 '16 at 7:36
  • $\begingroup$ OK, then it might help if you add to your question which cases you allready considered and what did you achive/where do you get stuck. $\endgroup$ – Qaswed Jun 6 '16 at 8:16

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