Formula for matrix product

Question

Consider a sequence of matrices $A_1,\cdots, A_n$. Is there a nice way in which you could write $$(I-A_1)(I-A_2)\cdots (I-A_n)$$ in terms of some summation? For the first few terms we would get $$(I-A_1-A_2+A_1A_2)(I-A_3)\cdots (I-A_n)=(I-A_1-A_2+A_1A_2-A_3+A_1A_3+A_2A_3-A_1A_2A_3)(I-A_4)\cdots(I-A_n)$$ but I can't seem to see any pattern emerging. My guess would be someting like the following. $$(I-A_1)(I-A_2)\cdots (I-A_n)=I-\sum_{k=1}^nA_k +\sum_{(i,j)\in \pi}(-1)^{n+1} A_iA_j+(-1)^{n+1}\prod_{k=1}^nA_k$$ but I'm not too sure what I would be missing here. I am fairly certain what I have here doesn't work. Here $\pi$ denotes some permutation of $\{1,\cdots,n\}$

Answer 1

In the following we use the notation $[n]:=\{1,2,\ldots,n\}$.

We obtain \begin{align*} \color{blue}{\prod_{k=1}^n\left(I-A_k\right)}&=\sum_{S\subseteq [n]}\prod_{j\in S}\left(-A_j\right)\tag{1}\\ &=\sum_{S\subseteq [n]}(-1)^{|S|}\prod_{j\in S}A_j\tag{2}\\ &\,\,=\color{blue}{\sum_{q=0}^n(-1)^q\sum_{{S\subseteq n}\atop{|S|=q}}\prod_{j\in S}A_j}\tag{3} \end{align*}

Comment:

• In (1) we select foreach subset $S\subset [n]$ the matrices $A_j,j\in\{S\}$.

• In (2) we factor out $-1$.

• In (3) we rearrange the sum according to increasing $|S|$.