When does $\det(A+B)=\det(A)+\det(B)$ hold?
I actually wonder if there is an easy answer for when $Per(A+B)=Per(A)+Per(B)$.
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let me assume $A$ is invertible, then you ask when $$\det(1+X)=1+\det X,\;\;X=A^{-1}B $$ so if $X$ has eigenvalues $x_i$, $i=1,2,\ldots n$, you would need $$\prod_{i}(1+x_i)=1+\prod_i x_i$$ basically you can take arbitrary values for $x_1,x_2,\ldots x_{n-1}$ and then the only requirement is that $$x_n=\frac{1-U}{U-V},\;\;U=\prod_{i=1}^{n-1}(1+x_i),\;\;V=\prod_{i=1}^{n-1}x_i$$
answered Jul 8, 2017 at 12:01
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5$\begingroup$ @Turbo, the given identity formula for $x_n$ is just a rearrangement of the desired identity, so it's necessary and sufficient. There is certainly a formula involving only the entries—just write out the determinants in terms of the entries—but it seems unlikely to be very interesting …. $\endgroup$– LSpiceCommented Jul 8, 2017 at 13:20
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2$\begingroup$ @Turbo, I'm no expert, but, since you are asking for equalities rather than, say, asymptotics of computing the determinant, I'm not sure how sparsity could help. (After all, for a fixed $n$, it's just as valid to say that a given matrix has $\mathrm O(n)$ as $\mathrm O(1)$ non-zero entries per row.) $\endgroup$– LSpiceCommented Jul 8, 2017 at 15:06
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$\begingroup$ @CarloBeenakker does it work for matrices over rings with no zero divisors? $\endgroup$– TurboCommented Jul 11, 2017 at 7:09
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2$\begingroup$ If you take the field of fractions, you can basically reduce the case of an integral domain to that of a field (you only have to check whether the expression for $x_n$ you get still lies in the ring). $\endgroup$ Commented Jul 11, 2017 at 8:15