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Let $A$ and $B$ be $n×n$ matrices. If we expand the determinant of $A+B$ as a sum over all permutations of $[\![1,n]\!]$ and all choices of whether the coefficient comes from $A$ or from $B$, this gives a sum with $2^n n!$ terms. We can rearrange it to give a sum over all minors of $A$ of this minor, times the "complementary" minor in $B$, times a sign.

It is a bit hard to explain so I made a program (in sage): https://pastebin.com/qxVRbQMG.

But this formula can be expressed more abstractly. We remark that the $k×k$ minors of $A$ are the coefficients of $\Lambda^k A$ (the exterior power) in the canonical basis. Using the perfect pairing $\Lambda^k E ⊗ \Lambda^{n-k} E → \Lambda^n E$, given an endomorphism $f$ of $\Lambda^k E$, we can define its adjoint $f^* : \Lambda^{n-k} E → \Lambda^{n-k} E$ relatively to this pairing. Given an endomorphism $f$ of $\Lambda^k E$ and an endomorphism $g$ of $\Lambda^{n-k} E$, we define $\langle f, g \rangle$ as $\operatorname{tr}(f^* g)$ (this is again a perfect pairing). Then, the formula translates as $$\det(A+B) = \sum_{k=0}^n \left\langle \Lambda^k A, \Lambda^{n-k} B \right\rangle\text{.}$$

Question: Is there an abstract proof (coordinate-free) of this identity? An intuition behind it? Can we generalize it if it helps?

amWhy
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Dabouliplop
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    Hint: There is a natural (in both $V$ and $W$) isomorphism $\wedge^n\left(V \oplus W\right) \cong \bigoplus\limits_{k=0}^n \left(\wedge^k V\right) \otimes \left(\wedge^{n-k} W\right)$ whenever $V$ and $W$ are two modules over a commutative ring. This isomorphism sends each $\left(v_1, w_1\right) \wedge \left(v_2, w_2\right) \wedge \cdots \wedge \left(v_n, w_n\right)$ to the element of $\bigoplus\limits_{k=0}^n \left(\wedge^k V\right) \otimes \left(\wedge^{n-k} W\right)$ whose $k$-th component (for each $k \in \left{0,1,\ldots,n\right}$) is ... – darij grinberg Mar 04 '19 at 21:04
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    ... the sum $\sum_\sigma \left(-1\right)^\sigma \left(v_{\sigma\left(1\right)} \wedge v_{\sigma\left(2\right)} \wedge \cdots \wedge v_{\sigma\left(k\right)} \right) \otimes \left(w_{\sigma\left(k+1\right)} \wedge w_{\sigma\left(k+2\right)} \wedge \cdots \wedge w_{\sigma\left(n\right)}\right)$ over all permutations $\sigma \in S_n$ satisfying $\sigma\left(1\right) < \sigma\left(2\right) < \cdots < \sigma\left(k\right)$ and $\sigma\left(k+1\right) < \sigma\left(k+2\right) < \cdots < \sigma\left(n\right)$. The inverse map is even easier to write down. – darij grinberg Mar 04 '19 at 21:08
  • @darijgrinberg Thank you for your answer! So we write $A+B$ as the composition of $V→V⊕V→V⊕V→V$, where the middle arrow is $A⊕B$. Then we apply the functor $Λ^n$, where $n=\dim(V)$. We can use your natural isomorphism and separate the endomorphism $Λ^nV$ as a sum over $k$ of endomorphisms. For each of these endomorphisms, we want its trace. This gives a bilinear mapping $\newcommand{\End}{\operatorname{End}}\End(Λ^kV)⊗\End(Λ^{n-k}V)→$, and we want to show that it coïncides with the map $Λ^kV^⊗Λ^kV⊗Λ^{n-k}V^⊗Λ^{n-k}V→Λ^nV^*⊗Λ^nV→$. We only need to check this on pure elements $f⊗x⊗g⊗y$. – Dabouliplop Mar 05 '19 at 00:03
  • Such an element is sent on the mapping that sends $v_1∧⋯∧v_n$ on $∑(-1)^f(v{(1)}∧⋯∧v_{(k)})x∧g(v_{(k+1)}∧⋯∧v_{(n)})y$. That is, $(f∧g)(v_1∧⋯∧v_n) x∧y$. Taking the trace of this, we get $(f∧g)(x∧y)$, what we wanted. Is it the thing you had in mind? – Dabouliplop Mar 05 '19 at 00:03
  • At this point, I neither remember what I had in mind nor fully get what you're doing (how do you get your bilinear mapping?). But probably I just should go to bed. – darij grinberg Mar 05 '19 at 04:35
  • @darijgrinberg Good night. Even if I'm not fully satisfied with this it is already good. By the way, I was looking at your website and found the same formula on your document "Notes on the combinatorial fundamentals of algebra", page 512 (section 6.32). I may post a self-answer to give the details, but later. – Dabouliplop Mar 05 '19 at 04:42
  • Yeah, I did give a proof there, but it's hardly the abstract argument you're looking for :) – darij grinberg Mar 05 '19 at 05:04
  • This can also be explained using the box product. – blargoner Nov 11 '23 at 03:39
  • @blargoner Thanks! However, I still don't think it is satisfying. There are two components to the proof:
    1. the decomposition $Λ^k(A+B)=∑_{p+q}Λ^pA☐Λ^qB$, which darij gave above with a coordinate-free proof working in any characteristic (not relying on factorials, with $f☐g$ defined as some composite $Λ^nV→Λ^pV⊗Λ^qV→Λ^pV⊗Λ^qV→Λ^nV$)
    2. the fact that $\operatorname{tr}(f☐g)=\langle f,g\rangle$, which I still don't know how to explain. Or maybe I knew and I don't remember now... anyway, if you have a complete answer, it is welcome
     – Dabouliplop Nov 12 '23 at 14:09
  • @Dabouliplop Yes, it's also possible to give a coordinate-free version using mixed exterior algebra (see here), which also explains the connection between trace and scalar product in terms of an isometry. But it would take me a bit of work to translate that into the formalism in your post, using a mixed Poincaré dual map. I just wanted to point out the connection for awareness. :) – blargoner Nov 12 '23 at 16:46
  • @blargoner Ok, that looks interesting, thanks. You indeed seem to show that $\operatorname{tr}(f☐g))=\operatorname{tr}(f^*g)$ but I'm not smart enough to check it. I'm only bothered by the fact that you divide by factorials and the myriad of ways of doing things, like the other coordinate-free definition of $☐$ using $(u⊗x)☐(v⊗y)=(u∧v)⊗(x∧y)$. What a mess, and even if I took the time to understand it, it would bring nothing in the end... oh well. – Dabouliplop Nov 14 '23 at 00:02

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