Why $\bigwedge^{d-1}A=\bigwedge^{d-1}B \Rightarrow A= \pm B$

Question

Let $V,W$ be $d$-dimensional vector spaces, and let $A,B \in \text{Hom}(V,W)$.

Consider the induced maps on the exterior algebras: $\bigwedge^{d-1}A,\bigwedge^{d-1}B :\Lambda_{d-1}(V) \to \Lambda_{d-1}(W)$.

Suppose that $\bigwedge^{d-1}A=\bigwedge^{d-1}B$, and that $A,B$ are invertible.

I want to prove that $A=\pm B$. (Note that this implies $A=B$ in the case $d$ is even).

The assumption implies $$\bigwedge^{d-1}(AB^{-1})=\text{Id}_{\Lambda_{d-1}(V)}.$$

Hence, the problem reduces to showing that for $S \in \text{GL}(V)$ ,$$\bigwedge^{d-1}S=\text{Id}_{\Lambda_{d-1}(V)} \Rightarrow S=\pm \text{Id}_V.$$

I can show this by introducing an inner product and orientation on $V$, but this seems "unnatural" to me.

Is there a proof which avoids this?

Answer 1

Let us consider the reduced version. We have $S\in \text{GL}(V)$ such that $$\bigwedge^{d-1}S=\text{id}.$$ In particular, this means that every $d-1$-dimensional subspace is invariant under $S$. On the other hand, for any $v,w\in V$ such that $w\not\in\langle v\rangle$, there is a $d-1$-dimensional subspace containing $v$ but not $w$. It thus follows that every element of $V$ is an eigenvector of $S$, and so, $S$ is multiplication by a scalar $\alpha$. Finally, we must have $\alpha^{d-1}=1$.

As pointed out by Gunnar, this does not necessarily mean that $\alpha=\pm1$ if we allow fields with more roots of unity.

Answer 2

If you choose a base of $V_d$ say $e_1,...e_d$, and declare $E_i=(-1)^i e_1 \wedge .. e_{i-1}\wedge e_i... \wedge e_n$ be a base of $\wedge^{d-1} V_d$. If $F$ is an endomorphism of $V_d$, with matrix $M$, then the matrix of $\wedge ^{d-1} f$ is the comatrix $\tilde M$ of $M$ and satisfies ${\tilde M}. M= \det M. Id$.

So we are reduced to prove that if two invertible $(d,d)$ matrices have $\det(M)^{-1}M= \det(N)^{-1} N$ then $M=N$.

Taking the determinant we see that $det(M)^{-d+1}=det (N)^{-d+1}$, and the result follows in the case the field of coefficient is $\bf R$. Note that over $\bf C$ the result is false.