# 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 :

$$\det(A+B+C)-\left[\det(A+B)+\det(A+C)+\det(B+C)\right]+\det(A)+\det(B)+\det(C)=0$$

• This is directly related to this MSE question – Somos Nov 14 '17 at 19:31
• @Somos: Thank you ! I will jump to it right now :) – Adren Nov 14 '17 at 19:47

Let me outline two other proofs. Let me first rename your $$m$$ and $$n$$ as $$n$$ and $$r$$, since I find it confusing when $$n$$ is not the size of the square matrices involved. So you are claiming the following:

Theorem 1. Let $$\mathbb{K}$$ be a commutative ring. Let $$n\in\mathbb{N}$$ and $$r\in\mathbb{N}$$ be such that $$n. Let $$A_{1},A_{2},\ldots,A_{r}$$ be $$n\times n$$-matrices over $$\mathbb{K}$$. Then, $$$$\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\det\left( \sum\limits_{i\in I}A_{i}\right) =0.$$$$

Notice that I've snuck in one more little change into your formula: I've added the addend for $$I=\varnothing$$. This addend usually doesn't contribute much, because $$\det\left( \sum\limits_{i\in\varnothing}A_{i}\right) =\det\left( 0_{n\times n}\right)$$ is usually $$0$$... unless $$n=0$$, in which case it contributes $$\det\left( 0_{0\times0}\right) =1$$ (keep in mind that there is only one $$0\times0$$-matrix and its determinant is $$1$$), and the whole equality fails if this addend is missing.

A first proof of Theorem 1 appears in (the solution to) Exercise 6.53 in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. (To obtain Theorem 1 from this exercise, set $$G=\left\{ 1,2,\ldots,r\right\}$$.) The main idea of this proof is that Theorem 1 holds not only for determinants, but also for each of the $$n!$$ products that make up the determinant (assuming that you define the determinant of an $$n\times n$$-matrix as a sum over the $$n!$$ permutations); this is proven by interchanging summation signs and exploiting discrete "destructive interference" (i.e., the fact that if $$G$$ is a finite set and $$R$$ is a subset of $$G$$, then $$\sum\limits_{\substack{I\subseteq G;\\R\subseteq I}}\left( -1\right) ^{\left\vert I\right\vert }= \begin{cases} 1, & \text{if }R=G;\\ 0, & \text{if }R\neq G \end{cases}$$).

Let me now sketch a second proof of Theorem 1, which shows that it isn't really about determinants. It is about finite differences, in a slightly more general context than they are usually studied.

Let $$M$$ be any $$\mathbb{K}$$-module. The dual $$\mathbb{K}$$-module $$M^{\vee }=\operatorname{Hom}_{\mathbb{K}}\left( M,\mathbb{K}\right)$$ of $$M$$ consists of all $$\mathbb{K}$$-linear maps $$M\rightarrow\mathbb{K}$$. Thus, $$M^{\vee}$$ is a $$\mathbb{K}$$-submodule of the $$\mathbb{K}$$-module $$\mathbb{K}^{M}$$ of all maps $$M\rightarrow\mathbb{K}$$. The $$\mathbb{K}$$-module $$\mathbb{K}^{M}$$ becomes a commutative $$\mathbb{K}$$-algebra (we just define multiplication to be pointwise, i.e., the product $$fg$$ of two maps $$f,g:M\rightarrow\mathbb{K}$$ sends each $$m\in M$$ to $$f\left( m\right) g\left( m\right) \in\mathbb{K}$$).

For any $$d\in\mathbb{N}$$, we let $$M^{\vee d}$$ be the $$\mathbb{K}$$-linear span of all elements of $$\mathbb{K}^{M}$$ of the form $$f_{1}f_{2}\cdots f_{d}$$ for $$f_{1},f_{2},\ldots,f_{d}\in M^{\vee}$$. (For $$d=0$$, the only such element is the empty product $$1$$, so $$M^{\vee0}$$ consists of the constant maps $$M\rightarrow\mathbb{K}$$. Notice also that $$M^{\vee1}=M^{\vee}$$.) The elements of $$M^{\vee d}$$ are called homogeneous polynomial functions of degree $$d$$ on $$M$$. The underlying idea is that if $$M$$ is a free $$\mathbb{K}$$-module with a given basis, then the elements of $$M^{\vee d}$$ are the maps $$M\rightarrow \mathbb{K}$$ that can be expressed as polynomials of the coordinate functions with respect to this basis; but the $$\mathbb{K}$$-module $$M^{\vee d}$$ makes perfect sense whether or not $$M$$ is free.

We also set $$M^{\vee d}=0$$ (the zero $$\mathbb{K}$$-submodule of $$\mathbb{K} ^{M}$$) for $$d<0$$.

For each $$d \in \mathbb{Z}$$, we define a $$\mathbb{K}$$-submodule $$M^{\vee \leq d}$$ of $$\mathbb{K}^M$$ by $$$$M^{\vee \leq d} = \sum\limits_{i \leq d} M^{\vee i} .$$$$ The elements of $$M^{\vee \leq d}$$ are called (inhomogeneous) polynomial functions of degree $$\leq d$$ on $$M$$. The submodules $$M^{\vee \leq d}$$ satisfy $$$$M^{\vee \leq d} M^{\vee \leq e} \subseteq M^{\vee \leq \left(d+e\right)}$$$$ for any integers $$d$$ and $$e$$.

For any $$x\in M$$, we define the $$\mathbb{K}$$-linear map $$S_{x}:\mathbb{K} ^{M}\rightarrow\mathbb{K}^{M}$$ by setting $$$$\left( S_{x}f\right) \left( m\right) =f\left( m+x\right) \qquad\text{for each }m\in M\text{ and }f\in\mathbb{K}^{M}.$$$$ This map $$S_{x}$$ is called a shift operator. It is an endomorphism of the $$\mathbb{K}$$-algebra $$\mathbb{K}^{M}$$ and preserves all the $$\mathbb{K}$$-submodules $$M^{\vee \leq d}$$ (for all $$d\in\mathbb{Z}$$).

Moreover, for any $$x\in M$$, we define the $$\mathbb{K}$$-linear map $$\Delta _{x}:\mathbb{K}^{M}\rightarrow\mathbb{K}^{M}$$ by $$\Delta_{x} =\operatorname*{id}-S_{x}$$. Hence, $$$$\left( \Delta_{x}f\right) \left( m\right) =f\left( m\right) -f\left( m+x\right) \qquad\text{for each }m\in M\text{ and }f\in\mathbb{K}^{M}.$$$$ This map $$\Delta_{x}$$ is called a difference operator. The following crucial fact shows that it "decrements the degree" of a polynomial function, similarly to how differentiation decrements the degree of a polynomial:

Lemma 2. Let $$x \in M$$. Then, $$\Delta_{x}M^{\vee d}\subseteq M^{\vee \leq \left( d-1\right)}$$ for each $$d\in\mathbb{Z}$$.

[Let me sketch a proof of Lemma 2:

Lemma 2 clearly holds for $$d < 0$$ (since $$M^{\vee d} = 0$$ if $$d < 0$$). Hence, it remains to prove Lemma 2 for $$d \geq 0$$. We shall prove this by induction on $$d$$. The induction base is the case $$d = 0$$, which is easy to check (indeed, each $$f \in M^{\vee 0}$$ is a constant map, and thus satisfies $$\Delta_x f = 0$$; therefore, $$\Delta_{x}M^{\vee 0} = 0 \subseteq M^{\vee \leq \left( 0-1\right) }$$).

For the induction step, we fix some nonnegative integer $$e$$, and assume that Lemma 2 holds for $$d = e$$. We must then show that Lemma 2 holds for $$d = e+1$$.

We have assumed that Lemma 2 holds for $$d = e$$. In other words, we have $$\Delta_{x}M^{\vee e}\subseteq M^{\vee \leq \left( e-1\right)}$$.

Our goal is to show that Lemma 2 holds for $$d = e+1$$. In other words, our goal is to show that $$\Delta_{x}M^{\vee \left(e+1\right)}\subseteq M^{\vee \leq e}$$.

But the $$\mathbb{K}$$-module $$M^{\vee \left(e+1\right)}$$ is spanned by maps of the form $$fg$$ with $$f\in M^{\vee e}$$ and $$g\in M^{\vee}$$ (since it is spanned by products of the form $$f_1 f_2 \cdots f_{e+1}$$ with $$f_1, f_2, \ldots, f_{e+1} \in M^{\vee}$$, but each such product can be rewritten in the form $$fg$$ with $$f = f_1 f_2 \cdots f_e \in M^{\vee e}$$ and $$g = f_{e+1} \in M^{\vee}$$). Hence, it suffices to show that $$\Delta_x \left( fg \right) \in M^{\vee \leq e}$$ for each $$f\in M^{\vee e}$$ and $$g\in M^{\vee}$$.

Let us first notice that if $$g \in M^{\vee}$$ is arbitrary, then $$\Delta_x g$$ is the constant map whose value is $$- g\left(x\right)$$ (because each $$m \in M$$ satisfies $$$$\left(\Delta_x g\right) \left(m\right) = g\left(m\right) - \underbrace{g\left(m+x\right)}_{\substack{=g\left(m\right) + g\left(x\right)\\ \text{(since }g \text{ is } \mathbb{K}\text{-linear)}}} = g\left(m\right) - \left(g\left(m\right) + g\left(x\right)\right) = - g\left(x\right)$$$$ ), and thus belongs to $$M^{\vee 0}$$. In other words, $$\Delta_x M^{\vee} \subseteq M^{\vee 0}$$.

For each $$f \in \mathbb{K}^M$$ and $$g \in \mathbb{K}^M$$, we have \begin{align*} \Delta_{x}\left( fg\right) & =\left( \operatorname*{id}-S_{x}\right) \left( fg\right) \qquad\left( \text{since }\Delta_{x}=\operatorname*{id} -S_{x}\right) \\ & =fg-\underbrace{S_{x}\left( fg\right) }_{\substack{=\left( S_{x}f\right) \left( S_{x}g\right) \\\text{(since }S_{x}\text{ is an endomorphism} \\\text{of the }\mathbb{K}\text{-algebra }\mathbb{K}^{M}\text{)}}}\\ & =fg-\left( S_{x}f\right) \left( S_{x}g\right) =\underbrace{\left( f-S_{x}f\right) }_{=\left( \operatorname*{id}-S_{x}\right) f}g+\left( S_{x}f\right) \underbrace{\left( x-S_{x}g\right) }_{=\left( \operatorname*{id}-S_{x}\right) g}\\ & =\left( \underbrace{\left( \operatorname*{id}-S_{x}\right) }_{=\Delta _{x}}f\right) g+\left( S_{x}f\right) \left( \underbrace{\left( \operatorname*{id}-S_{x}\right) }_{=\Delta_{x}}g\right) \\ & =\left( \Delta_{x}f\right) g+\left( \underbrace{S_{x}}_{\substack{=\operatorname*{id}-\Delta_{x}\\ \text{(since }\Delta _{x}=\operatorname*{id}-S_{x}\text{)}}}f\right) \left( \Delta_{x}g\right) \\ & =\left( \Delta_{x}f\right) g+\underbrace{\left( \left( \operatorname*{id}-\Delta_{x}\right) f\right) }_{=f-\Delta_{x}f}\left( \Delta_{x}g\right) \\ & =\left( \Delta_{x}f\right) g+\left( f-\Delta_{x}f\right) \left( \Delta_{x}g\right) \\ & =\left( \Delta_{x}f\right) g+f\left( \Delta_{x}g\right) -\left( \Delta_{x}f\right) \left( \Delta_{x}g\right) . \end{align*} Hence, for each $$f\in M^{\vee e}$$ and $$g\in M^{\vee}$$, we have \begin{align*} \Delta_{x}\left( fg\right) & =\left( \Delta_{x}\underbrace{f}_{\in M^{\vee e}}\right) \underbrace{g}_{\in M^{\vee}}+\underbrace{f}_{\in M^{\vee e}}\left( \Delta_{x}\underbrace{g}_{\in M^{\vee}}\right) -\left( \Delta _{x}\underbrace{f}_{\in M^{\vee e}}\right) \left( \Delta_{x}\underbrace{g}_{\in M^{\vee}}\right) \\ & \in\underbrace{\left( \Delta_{x}M^{\vee e}\right) }_{\subseteq M^{\vee \leq\left( e-1\right) }}M^{\vee}+M^{\vee e}\underbrace{\left( \Delta _{x}M^{\vee}\right) }_{\subseteq M^{\vee0}}-\underbrace{\left( \Delta _{x}M^{\vee e}\right) }_{\subseteq M^{\vee\leq\left( e-1\right) } }\underbrace{\left( \Delta_{x}M^{\vee}\right) }_{\subseteq M^{\vee0}}\\ & \subseteq\underbrace{M^{\vee\leq\left( e-1\right) }M^{\vee}}_{\subseteq M^{\vee\leq e}}+\underbrace{M^{\vee e}M^{\vee0}}_{\subseteq M^{\vee e}\subseteq M^{\vee\leq e}}-\underbrace{M^{\vee\leq\left( e-1\right) }M^{\vee0}}_{\subseteq M^{\vee\leq\left( e-1\right) }\subseteq M^{\vee\leq e}}\\ & \subseteq M^{\vee\leq e}+M^{\vee\leq e}-M^{\vee\leq e}\subseteq M^{\vee\leq e}. \end{align*} This proves that $$\Delta_{x}\left( M^{\vee\left( e+1\right) }\right) \subseteq M^{\vee\leq e}$$, as we intended to prove.

Thus, the induction step is complete, and Lemma 2 is proven.]

The following fact follows by induction using Lemma 2:

Corollary 3. Let $$r\in\mathbb{N}$$. Let $$x_{1},x_{2},\ldots,x_{r}$$ be $$r$$ elements of $$M$$. Then, $$$$\Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r}}M^{\vee d}\subseteq M^{\vee \leq \left( d-r\right) }$$$$ for each $$d\in\mathbb{Z}$$.

And as a consequence of this, we obtain the following:

Corollary 4. Let $$r\in\mathbb{N}$$. Let $$x_{1},x_{2},\ldots,x_{r}$$ be $$r$$ elements of $$M$$. Then, $$$$\Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r}}M^{\vee d}=0$$$$ for each $$d\in\mathbb{Z}$$ satisfying $$d.

[In fact, Corollary 4 follows immediately from Corollary 3, because $$d implies $$M^{\vee \leq \left( d-r\right) }=0$$.]

To make use of Corollary 4, we want a more-or-less explicit expression for how $$\Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r}}$$ acts on maps in $$\mathbb{K}^{M}$$. This is the following fact:

Proposition 5. Let $$r\in\mathbb{N}$$. Let $$x_{1},x_{2},\ldots,x_{r}$$ be $$r$$ elements of $$M$$. Then, $$$$\left( \Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r}}f\right) \left( m\right) =\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }f\left( m+\sum\limits_{i\in I}x_{i}\right) \qquad\text{for each }m\in M\text{ and }f\in\mathbb{K}^{M}.$$$$

[Proposition 5 can be proven by induction over $$r$$, where the induction step involves splitting the sum on the right hand side into the part with the $$I$$ that contain $$r$$ and the part with the $$I$$ that don't. But there is also a slicker argument, which needs some preparation. The maps $$S_{x}\in \operatorname{End}_{\mathbb{K}}\left( \mathbb{K}^{M}\right)$$ for different elements $$x\in M$$ commute; better yet, they satisfy the multiplication rule $$S_{x}S_{y}=S_{x+y}$$ (as can be checked immediately). Hence, by induction over $$\left\vert I\right\vert$$, we conclude that if $$I$$ is any finite set, and if $$x_{i}$$ is an element of $$M$$ for each $$i\in I$$, then $$$$\prod\limits_{i\in I}S_{x_{i}}=S_{\sum\limits_{i\in I}x_{i}} \qquad \text{in the ring } \operatorname{End}_{\mathbb{K}} \left(\mathbb{K}^M\right) .$$$$ I shall refer to this fact as the S-multiplication rule.

Now, let us prove Proposition 5. Let $$x_{1},x_{2},\ldots,x_{r}$$ be $$r$$ elements of $$M$$. Recall the well-known formula $$$$\prod\limits_{i\in\left\{ 1,2,\ldots,r\right\} }\left( 1-a_{i}\right) =\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\prod\limits_{i\in I}a_{i},$$$$ which holds whenever $$a_{1},a_{2},\ldots,a_{r}$$ are commuting elements of some ring. Applying this formula to $$a_{i}=S_{x_{i}}$$, we obtain $$$$\prod\limits_{i\in\left\{ 1,2,\ldots,r\right\} }\left( \operatorname*{id} -S_{x_{i}}\right) =\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\prod\limits_{i\in I}S_{x_{i}}$$$$ (since $$S_{x_{1}},S_{x_{2}},\ldots,S_{x_{r}}$$ are commuting elements of the ring $$\operatorname{End}_{\mathbb{K}}\left( \mathbb{K}^{M}\right)$$). Thus, \begin{align*} \Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r}} & =\prod\limits_{i\in\left\{ 1,2,\ldots,r\right\} }\underbrace{\Delta_{x_{i}}} _{\substack{=\operatorname*{id}-S_{x_{i}}\\\text{(by the definition of } \Delta_{x_{i}}\text{)}}}=\prod\limits_{i\in\left\{ 1,2,\ldots,r\right\} }\left( \operatorname*{id}-S_{x_{i}}\right) \\ & =\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\underbrace{\prod\limits_{i\in I}S_{x_{i}}} _{\substack{=S_{\sum\limits_{i\in I}x_{i}}\\\text{(by the S-multiplication rule)} }}=\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }S_{\sum\limits_{i\in I}x_{i}}. \end{align*} Hence, for each $$m\in M$$ and $$f\in\mathbb{K}^{M}$$, we obtain \begin{align*} & \left( \Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r}}f\right) \left( m\right) \\ & =\left( \sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }S_{\sum\limits_{i\in I}x_{i}}f\right) \left( m\right) \\ & =\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\underbrace{\left( S_{\sum\limits_{i\in I}x_{i}}f\right) \left( m\right) }_{\substack{=f\left( m+\sum\limits_{i\in I}x_{i}\right) \\\text{(by the definition of the shift operators)}}}\\ & =\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }f\left( m+\sum\limits_{i\in I}x_{i}\right) . \end{align*} Thus, Proposition 5 is proven.]

We can now combine Corollary 4 with Proposition 5 and obtain the following:

Corollary 6. Let $$x_{1},x_{2},\ldots,x_{r}$$ be $$r$$ elements of $$M$$. Let $$d\in\mathbb{Z}$$ be such that $$d. Let $$f\in M^{\vee d}$$ and $$m\in M$$. Then, $$$$\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }f\left( m+\sum\limits_{i\in I}x_{i}\right) =0.$$$$

[Indeed, Corollary 6 follows from the computation \begin{align*} & \sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }f\left( m+\sum\limits_{i\in I}x_{i}\right) \\ & =\underbrace{\left( \Delta_{x_{1}}\Delta_{x_{2}}\cdots\Delta_{x_{r} }f\right) }_{\substack{=0\\\text{(by Corollary 4, since } f \in M^{\vee d} \text{)}}}\left( m\right) \qquad\left( \text{by Proposition 5}\right) \\ & =0. \end{align*} ]

Finally, let us prove Theorem 1. The matrix ring $$\mathbb{K}^{n\times n}$$ is a $$\mathbb{K}$$-module. Let $$M$$ be this $$\mathbb{K}$$-module $$\mathbb{K}^{n\times n}$$. For each $$i,j\in\left\{ 1,2,\ldots,n\right\}$$, we let $$x_{i,j}$$ be the map $$M\rightarrow\mathbb{K}$$ that sends each matrix $$M$$ to its $$\left( i,j\right)$$-th entry; this map $$x_{i,j}$$ is $$\mathbb{K}$$-linear and thus belongs to $$M^{\vee}$$.

It is easy to see that the map $$\det:\mathbb{K}^{n\times n}\rightarrow \mathbb{K}$$ (sending each $$n\times n$$-matrix to its determinant) is a homogeneous polynomial function of degree $$n$$ on $$M$$; indeed, it can be represented in the commutative $$\mathbb{K}$$-algebra $$\mathbb{K}^M$$ as $$$$\det=\sum\limits_{\sigma\in S_{n}}\left( -1\right) ^{\sigma}x_{1,\sigma\left( 1\right) }x_{2,\sigma\left( 2\right) }\cdots x_{n,\sigma\left( n\right) },$$$$ where $$S_{n}$$ is the $$n$$-th symmetric group, and where $$\left( -1\right) ^{\sigma}$$ denotes the sign of a permutation $$\sigma$$. In other words, $$\det\in M^{\vee n}$$. Hence, Corollary 6 (applied to $$x_{i}=A_{i}$$, $$d=n$$, $$f=\det$$ and $$m=0$$) yields $$$$\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\det\left( 0+\sum\limits_{i\in I}A_{i}\right) =0.$$$$ In other words, $$$$\sum\limits_{I\subseteq\left\{ 1,2,\ldots,r\right\} }\left( -1\right) ^{\left\vert I\right\vert }\det\left( \sum\limits_{i\in I}A_{i}\right) =0.$$$$ This proves Theorem 1. $$\blacksquare$$

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 very beautiful result and very beautiful proof! – Jair Taylor Nov 15 '17 at 18:22

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:

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

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