Determinant of a sum of matrices I would like to know if the following formula is well known and get some references for it.
I don't know yet how to prove it (and I am working on it), but I am quite sure of its validity, after having performed a few symbolic computations with Maple.
Given $n$ square matrices $A_1,\ldots,A_n$ of size $m<n$ :

$$\sum_{p=1}^n(-1)^p\sum_{1\leqslant i_1<\cdots<i_p\leqslant n}\det(A_{i_1}+\cdots+A_{i_p})=0$$

For example, if $A,B,C$ are three $2\times2$ matrices, then :
$$\det(A+B+C)-\left[\det(A+B)+\det(A+C)+\det(B+C)\right]+\det(A)+\det(B)+\det(C)=0$$
 A: Given integers $n > m > 0$, let $[n]$ be a short hand for the set $\{1,\ldots,n\}$.
For any $t \in \mathbb{R}$ and $x_1, \ldots, x_n \in \mathbb{C}$, we have the identity
$$\prod_{k=1}^n (1 - e^{tx_k}) = \sum_{P \subset [n]} (-1)^{|P|} e^{t\sum_{k\in P} x_k}$$
Treat both sides as function of $t$. Expand against $t$, one notice on LHS, coefficients in front of $t^k$ vanishes whenever $k < n$. 
By comparing coefficients of $t^m$, we obtain:
$$ 0 = \sum_{P\subset [n]} (-1)^{|P|} \left(\sum_{k\in P} x_k\right)^m\tag{*1}$$
Notice RHS is a polynomial function in $x_1,\ldots,x_n$ with integer coefficients. Since it evaluates to $0$ for all $(x_1,\ldots,x_n) \in \mathbb{C}^n$, it is valid as a polynomial identity in $n$ indeterminates with integer coefficients. As a corollary, it is valid as an algebraic identity when $x_1, x_2, \ldots, x_n$ are elements taken from any commutative algebra.
Let $V$ be a vector space over $\mathbb{C}$ spanned by
elements $\eta_1, \ldots, \eta_m$ and $\bar{\eta}_1,\ldots,\bar{\eta}_m$.
Let $\Lambda^{e}(V) = \bigoplus_{k=0}^n \Lambda^{2k}(V)$ be the 'even' portion
of its exterior algebra. $\Lambda^{e}(V)$ itself is a commutative algebra.
For any $m \times m$ matrix $A$, let $\tilde{A} \in \Lambda^e(V)$ be the element defined by: 
$$A = (a_{ij}) \quad\longrightarrow\quad \tilde{A} = \sum_{i=1}^m\sum_{j=1}^m a_{ij}\bar{\eta}_i \wedge \eta_j$$
Notice the $m$-fold power of $\tilde{A}$ satisfies an interesting identity:
$$\tilde{A}^m = \underbrace{\tilde{A} \wedge \cdots \wedge \tilde{A}}_{m \text{ times}} = \det(A) \omega
\quad\text{ where }\quad
\omega = m!\, \bar{\eta}_1 \wedge \eta_1 \wedge \cdots \wedge \bar{\eta}_m \wedge \eta_m\tag{*2}$$
Given any $n$-tuple of matrices $A_1, \ldots, A_n \in M_{m\times m}(\mathbb{C})$, if we substitute $x_k$ in $(*1)$ by $\tilde{A}_k$ and apply $(*2)$, we find
$$
\sum_{P\subset [n]} (-1)^{|P|} \left(\sum_{k\in P} \tilde{A}_k\right)^m
= \sum_{P\subset [n]} (-1)^{|P|} \det\left(\sum_{k\in P} A_k\right)\omega
= 0
$$
Extracting the coefficient in front of $\omega$, the desired identity follows:
$$\sum_{P\subset [n]} (-1)^{|P|} \det\left(\sum_{k\in P} A_k\right) = 0$$
A: HINT:
The determinant of an $n\times n$ matrix is a form of degree $n$. Forms come from multilinear forms. 
Consider $M$ an abelian group. For $a \in M$, denote by $a^{[n]}$ the element $a\otimes a \otimes \ldots \otimes a\in M^{\otimes n}$. Let now $a_i\in M$, $i \in I$, finitely many elements in $M$. Let's try to find 
$$\sum_{J\subset I}(-1)^{|I|-|J|}(\sum_{i \in J} a_i)^{[n]}$$
Consider a product $a_{i_1}\otimes \ldots \otimes a_{i_n}$. It appears in the above sum with the coefficient
$$\sum_{J\subset K \subset I}(-1)^{|I| - |J|}$$ where $J=\{i_1, \ldots, i_n \}$. This is $0$ for $J\ne I$ and $1$ for $J=I$. ( a Möbius function)
Therefore
$$\sum_{J\subset I}(-1)^{|I|-|J|}(\sum_{i \in J} a_i)^{[n]}=\sum_{\phi\colon \{1,\ldots n\}\to I,\phi\  \text{surjective}}a_{\phi(1)}\otimes \ldots a_{\phi(n)}$$
Particular cases:


*

*$|I|>n$, we get $0$, the result desired. 

*$|I|=n$, we get $\sum_{\phi\colon \{1,\ldots n\}\to I,\phi\  \text{bijective}}a_{\phi(1)}\otimes \ldots a_{\phi(n)}$
