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When does $\det(A+B)=\det(A)+\det(B)$ hold?

Is there necessary and sufficient condition?

Turbo
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    Related: Does det(A+B)=det(A)+det(B) hold?. – dxiv Oct 24 '16 at 22:36
  • @dxiv may be but only one post there (by PolyaPal) is relevant. – Turbo Oct 24 '16 at 22:36
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    It is related (and I didn't say duplicate). One of the answers appears to provide a certain sufficient condition. – dxiv Oct 24 '16 at 22:40
  • @dxiv not that is not relevant (only Polayapal's is relevant). – Turbo Oct 24 '16 at 22:41
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    An obvious case is: When the dimension is $1$ (that is, the matrices consist of just one number each). Another obvious case: If one of the two matrices is the zero matrix. A less obvious case: $A$ and $B$ are upper triangular matrices, and one of them has only zeros on the diagonal. I don't think those cases are exhaustive. – celtschk Oct 24 '16 at 22:53
  • Think of $\det(A)$ as the (oriented volume) of the set $$ A ( \mbox{the unit cube} ). $$ So, start in 2 dimensions and try to come up with transformations $A$ and $B$ such that the equality holds; i.e., specifically, such that the vector sum of the sets $A ( \mbox{the unit cube} )$ and $B ( \mbox{the unit cube} )$ gives a set whose volume equals $$ \mbox{vol}[A ( \mbox{the unit cube} ) ] + \mbox{vol}[B ( \mbox{the unit cube} ) ]. $$ My intuition is that this additivity is rather an exception than the rule. – user8960 Oct 25 '16 at 00:37

1 Answers1

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IT IS NOT AN ANSWER.

It is a long comment to show that problem in arbitrary dimension is far from trivial. If we set simplified case: $B = I$, then: $$ n = 2: \det(I + A) = 1 + \det(A) + Tr(A) $$$$ n = 3: \det(I + A) = 1 + \det(A) + Tr(A) + \frac{Tr^2(A) - Tr(A^2)}{2} $$$$ n = 4: \det(I + A) = 1 + \det(A) + Tr(A) + \frac{Tr^2(A) - Tr(A^2)}{2} + \frac{Tr^3(A) - 3Tr(A)Tr(A^2) + 2Tr(A^3)}{6} $$ General sequence comes from relation between determinant and exponent of the trace of the matrix logarithm. I doubt that there is a trivial solution in general case.

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