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I was reading the book Introduction to Random Matrices by Anderson et al and I think that the authors are using the following characterization of $\det(I-A)$ in Lemma $3.2.4$ in the book.

$$\det(I-A)=1+\sum_{k=1}^{n}\,\sum_{1\leq\nu(1)<...<\nu(k)\leq n}(-1)^{k}\det(a_{\nu(i)\nu(j)})_{i,j=1}^{k}$$

But I have never seen such a characterization and neither can I find any resources on this. I am unable to prove it on my own and it should involve a fair bit of Alternating algebra to prove it So if anyone can help me out by suggesting a reference or by telling me more about this then I'll be grateful. Also please tell me if this is at all true.

Blitzkrieg
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    Essentially, when the $i$ in not in the range, $\nu(1),\dots,\nu(k),$ we are taking the $1$ from the $i$th diagonal element. So the set of permutations which fix those $n-k$ $i$ values will contribute $\pm$ the determinant in the sum when we just take the $1$ terms in the product. The sign argument is the messy part. – Thomas Andrews Mar 26 '23 at 13:48
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    So if you write the determinant a $$\sum_{\pi\in S_n} \operatorname{sgn}(\pi)\prod_i(\delta_{i\pi(i)}-a_{\pi_i})$$ we get all the terms of $(-1)^n\det A,$ plus additional terms $$\sum_{J\subseteq [n]},\sum_{\pi\mid \pi j=j\forall j\in J} \operatorname{sgn}(\pi)\prod_{i\notin J} (-a_{i\pi(i)})$$ But the sign of $\pi$ is the same if treated as a permutation on $[n]={1,\dots,n}$ or restricted to $[n]\setminus J,$ since it is the identity on $J.$ – Thomas Andrews Mar 26 '23 at 14:04
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    Essentially, the $J\subseteq [n]$ is the set of indices of diagonals that are contributing $1$ rather than $a_{jj}.$ So the inner sum is $(-1)^{n-|J|}\det(a_{i,j})_{i,j\notin J}.$ – Thomas Andrews Mar 26 '23 at 14:05
  • @ThomasAndrews Perfect. I was looking precisely for an intuitive explanation. Unfortunately the question is closed so it is not an answer otherwise I'd have accepted it(and upvoted). – Blitzkrieg Mar 26 '23 at 14:35

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