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I have an equality which I am struggling to grasp: in an article the author says that $$ \det(I-A)=\sum_k (-1)^k \operatorname{tr}(\wedge^k A), $$ where $\wedge^k A$ is the map induced by $A$ on the $k$-th degree of the alternating algebra $\bigwedge^{\!*} V$. ($V$ has finite dimension, so the summands definitively vanish).

I wanted to ask if there is a representation of $\wedge^k A$, and how I can prove the equality. I want to apologize in advance, I know that it has to be some simple multilinear algebra stuff, but I don't really know much about it (not even a good reference book, which would be greatly appreciated).

I want to thank in advance everybody who will answer this.

Sammy Black
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    This post might help you: https://math.stackexchange.com/questions/1630815/cayley-hamilton-theorem-trace-of-exterior-power-form The OP there made a follow-up question in an attempt to prove the formula you are interested in. You may be able to extract a proof from that. –  Nov 19 '23 at 21:40
  • Thank you very much! –  Nov 19 '23 at 21:47
  • Perhaps, to simplify computations, you can do them in the algebraic closure of the given field, using Jordan normal form. – jg1896 Nov 19 '23 at 22:12
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    It's maybe a little easier to first prove $$\det(A+I)=\sum_k\mathop{\mathrm{tr}}(\wedge^k A)$$ to avoid dealing with one set of sign factors, then substitute $-A$ for $A$. One way of proving the latter is by applying Cauchy-Binet to the product of the block matrices $[A I]$ and $[I I]^T$. – blargoner Nov 20 '23 at 00:28

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Work over an algebraically closed field, so $A$ has generalized eigenvalues $\lambda_1,\dots,\lambda_n$. Then the generalized eigenvalues of $\wedge^kA$ are of the form $\lambda_{i_1}\cdots\lambda_{i_k}$ where $1\le i_1<\cdots<i_k\le n$. Now the equality is simply the classical identity $$\prod_{i=1}^n(1-\lambda_i)=\sum_{k=0}^n(-1)^k\sum_{i_1<\cdots<i_k}\lambda_{i_1}\cdots\lambda_{i_k}.$$

Kenta S
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