Determinant of a diagonal block matrix using exterior algebra

Question

Let $V$ and $U$ be two finite-dimensional vector spaces of dimensions $n$ and $m$ respectively. For $f\in \mathrm{End}\,V$ we define a multilinear, anti-symmetric function: $$V\times ...\times V\to \Lambda^nV,~~(v_1, v_2,\dots, v_n)\mapsto fv_1\wedge fv_2\wedge\dots\wedge fv_n$$ what corresponds to a linear function $\det f\colon \Lambda^nV\to\Lambda^nV$.

Now I would like to prove that $\det f\oplus g = \det f\cdot \det g$ (i. e. the determinant of a block diagonal matrix is the product of the determinants of blocks).

Naively, if $v_1, \dots, v_n$ and $u_1,\dots, u_m$ are bases of $V$ and $U$, we merge them into a basis of $V\oplus U$ and: $$ (f\oplus g)v_1\wedge \dots (f\oplus g)v_n\wedge (f\oplus g)u_1\wedge (f\oplus g)u_m = \\=(fv_1\wedge \dots\wedge fv_n) \wedge (gu_1\wedge\dots\wedge g u_m) =\\ =\det f\cdot \det g \cdot v_1\wedge\dots\wedge v_n \wedge u_1\wedge\dots\wedge u_m$$ However, I can't bracket out these two terms without some kind of identification between $\Lambda^{n+m}(V\oplus U)$ and pressumably $\Lambda^nV\otimes \Lambda^mU$. How does one fix this rigorously?

Edit: From Exterior power "commutes" with direct sum we know that there is an isomorphism $\Lambda ^{n+m}(V\oplus U) \simeq \Lambda^nV\otimes \Lambda^mU$.

Answer 1

Let $i_V\colon V\to V\oplus U$ and $i_U\colon U \to V\oplus U$ be the two inclusions. Now if $v_1, \dots, v_n$ is a basis of $V$ and $u_1,\dots, u_m$ is a basis of $U$, then: $$i_Vv_1, \dots, i_Vv_n, i_Uu_1, \dots, i_Uu_m$$ is a basis of $V\oplus U$. From the definition of $f\oplus g$ we know that $(f\oplus g)i_Vv_i=i_Vfv_i$ and $(f\oplus g)i_Uu_i=i_Ugu_i$. Now define: $$x := (f\oplus g)i_Vv_1\wedge \dots\wedge (f\oplus g)i_Vv_n \wedge(f\oplus g)i_Uu_1\wedge\dots\wedge (f\oplus g)i_Uu_m = \\ =i_Vfv_1\wedge\dots\wedge i_Vfv_n\wedge i_Ugu_1\wedge\dots\wedge i_Ugu_m .$$

We have a linear isomorphism $\alpha\colon \Lambda^{n+m}(V\oplus U)\to \Lambda^nV\otimes \Lambda^mU$. Acting with it we get an expression: $$\alpha(x) = (fv_1\wedge\dots\wedge fv_n)\otimes(gu_1\wedge\dots\wedge gu_m) \\ = \det f\cdot \det g\cdot (v_1\wedge\dots \wedge v_n)\otimes (u_1\wedge\dots\wedge u_m)$$ Using its inverse: $$x=(\alpha^{-1}\alpha)(x) = \det f\cdot \det g\cdot i_Vv_1\wedge\dots\wedge i_Vv_n\wedge i_Uu_1\wedge\dots\wedge i_Uu_m ,$$ which proves the claim.

We can reduce clutter resulting from inclusions – the crucial step is to move the calculation from $\Lambda^{n+m}(V\oplus U)$ to $\Lambda^nV\otimes \Lambda^mU$, and then translate the results back.