Let $V$ be a finite-dimensional vector space over a field $K$ of dimension $r:= \dim_K(V)$ (let's say $K$ equals the real or complex numbers).
If $\varphi: V\to V$ is an endomorphism, we know that $\varphi$ can be naturally extended to act as a derivation on $\bigwedge^k V $ for each $1\leq k\leq r$ by $\widetilde\varphi_k\colon \bigwedge^k V \to \bigwedge^k V $ defined as
$$ \widetilde\varphi_k(v_1\wedge\cdots\wedge v_k):=\sum_{j=1}^k v_1\wedge\cdots\wedge \varphi (v_j)\wedge\cdots\wedge v_k. $$
Now, I want a formal explanation/proof of why does this extension acts as just the scalar multiplication by the trace of $\varphi$ on the top exterior power $\bigwedge^r V \cong K$. Namely, I want to prove that
$$(*)\qquad\qquad\widetilde\varphi_r(v_1\wedge\cdots\wedge v_r)=\mathrm{tr}(\varphi)v_1\wedge\cdots\wedge v_r .$$
This has appeared in my path in the context of differential geometry. Namely, if $E$ is a vector bundle of rank $r$ with connection $\nabla$ and curvature form $\Omega$, then it is asserted or implicitely used in several places that the curvature form of the determinant line bundle $\det E:=\bigwedge^r E$ is given by $\widetilde\Omega=\mathrm{tr}(\Omega)$ with respect to the natural connection $\widetilde\nabla$ on $\bigwedge^r E$, namely, the extension of $\nabla$ as a derivation: $\widetilde\nabla(\xi_1\wedge\cdots\wedge \xi_r):=\sum\limits_{j=1}^r \xi_1\wedge\cdots\wedge \nabla \xi_j\wedge\cdots\wedge \xi_r$.
By what we said above, this just amounts to a multilinear algebra problem, that is, $(*)$.
It is also worth noting that the identity $(*)$ is a sister identity of the more well-known identity
$$\bigwedge\nolimits^r \varphi(v_1\wedge\cdots\wedge v_r):= \varphi(v_1)\wedge\cdots\wedge\varphi(v_r)=\det(\varphi)v_1\wedge\cdots\wedge v_r,$$
and that give us a nice overpowered definition of both the determinant and the trace. Any hint or proof is appreciated.
Best Answer
Let's choose a basis $(e_1,\dots,e_r)$ for $V$ and write $\varphi(e_i) = \sum_{j=1}^r \varphi_i^j e_j$. Then $\operatorname{tr}(\varphi) = \sum_{j=1}^r \varphi^j_j$ and
$$ \tilde{\varphi}(e_1 \wedge \dots \wedge e_r) = \sum_{j=1}^r e_1 \wedge \dots \wedge e_{j-1} \wedge \varphi(e_j) \wedge e_{j+1} \wedge \dots \wedge e_r = \\ \sum_{j=1}^r e_1 \wedge \dots \wedge e_{j-1} \wedge \varphi_j^k e_k \wedge e_{j+1} \wedge \dots \wedge e_r = \sum_{j=1}^r \varphi_j^j e_1 \wedge \dots \wedge e_r = \operatorname{tr}(\varphi) e_1 \wedge \dots \wedge e_r $$
where we used the fact that $e_1 \wedge \dots \wedge e_{j-1} \wedge e_k \wedge e_{j+1} \wedge e_r = 0$ unless $k = j$.