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.
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 $$\prod_{i=1}^n(1-\lambda_i)=\sum_{k=0}^n(-1)^k\sum_{i_1<\cdots<i_k}\lambda_{i_1}\cdots\lambda_{i_k}.$$