Why does a vector bundle have the same first Chern class as its determinant bundle?

Question

Let $A$ be a $2n$-dimensional complex vector bundle and $\det A=\Lambda^{2n}(A)$

How is $c_{1}(\det A)=c_{1}(A)$? Here $c_{1}$ means first Chern class.

Answer 1

By the splitting principle, it suffices to prove this in the case that $A$ is a direct sum of line bundles $L_1\oplus\dots\oplus L_n$. In that case, we have $$c_1(A)=c_1(L_1)+\dots+c_1(L_n)$$ by the Whitney sum formula, and on the other hand $\det A\cong L_1\otimes \dots\otimes L_n$ so $$c_1(\det A)=c_1(L_1\otimes\dots \otimes L_n)=c_1(L_1)+\dots+c_1(L_n).$$