Abtract explanation for $\det(A+B) = ∑_{k=0}^n \langle \Lambda^k A, \Lambda^{n-k} B \rangle$

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 $$\Lambda^{n-k} B$$, we define $$\langle f, g \rangle$$ as $$\operatorname{tr}(f^* g)$$ (this is again a perfect pairing). Then, the formula translates as $$\det(A+B) = ∑_{k=0}^n \langle \Lambda^k A, \Lambda^{n-k} B \rangle\text{.}$$

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

• 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 4 at 21:04
• ... 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 4 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 $\newcommand{\End}{\operatorname{End}}\End(Λ^kV)⊗\End(Λ^{n-k}V)→𝕂$, and we want to show that it coïncides with the map $Λ^kV^*⊗Λ^kV⊗Λ^{n-k}V^*⊗Λ^{n-k}V→Λ^nV^*⊗Λ^nV→𝕂$. We only need to check this on pure elements $f⊗x⊗g⊗y$. – Idéophage Mar 5 at 0:03
• Such an element is sent on the mapping that sends $v_1∧⋯∧v_n$ on $∑_𝜎(-1)^𝜎f(v_{𝜎(1)}∧⋯∧v_{𝜎(k)})x∧g(v_{𝜎(k+1)}∧⋯∧v_{𝜎(n)})y$. 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? – Idéophage Mar 5 at 0:03
• 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 5 at 4:35