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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?

Asaf Shachar
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  • I might be missing something obvious, but if you fix isomorphisms $\phi:V \to \Lambda^{d-1}V$, $\psi:W \to \Lambda^{d-1}W$, don't you have $A = \psi^{-1} (\Lambda^{d-1}A) \phi = \psi^{-1} (\Lambda^{d-1}B) \phi =B$? – Alex Provost Jul 07 '17 at 20:45
  • I think the question can be rephrased as "is there a proof that avoids arbitrary choice of isomorphism between $V$ and $\Lambda^{d-1}V$?" – Neal Jul 07 '17 at 20:46
  • Although it looks quite different, I believe the question to which you reduced your question is addressed here. Although the question concerns metric notions, the solution is just linear algebra. – Ted Shifrin Jul 07 '17 at 20:51
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    You're thinking of real vector spaces, right? Otherwise you get more solutions over fields with more roots of unity. – Gunnar Þór Magnússon Jul 07 '17 at 20:57
  • @AlexProvost Your approach is interesting. Can you say why there exist isomorphisms $\phi,\psi$ such that $A = \psi^{-1} (\Lambda^{d-1}A) \phi, \psi^{-1} (\Lambda^{d-1}B) \phi =B$ both hold? I can see why you can choose them so that one equality of the two will hold but not how can you achieve both. (You can choose an arbitrary one isomorphism of the two, and then the other one is forced). – Asaf Shachar Jul 08 '17 at 05:25
  • @Neal Thanks, this was exactly my intention. However, if there is a way to realize Alex's comment in some way, I would also be interested. (See my previous comment). – Asaf Shachar Jul 08 '17 at 05:26
  • @Ted Thanks, this does look similar to an argument I had in mind- but I have two caveats: (1) I think it requires introducing an inner product (and orientation) to use SVD. (2) I think it only works for diagonalizable maps $S$, since essentially you want to pass to $Sv_i=\sigma_i v_i$ (where $\sigma_i$ are the singular values, and then use the argument in the question you referred to. Actually this is the same argument I gave here. – Asaf Shachar Jul 08 '17 at 05:56
  • @AsafShachar Ah, I see what's wrong with my line of thought. The isomorphism $\psi$ I had in mind actually crucially depends on either $A$ or $B$, so there is no way to make both squares commute with a single $\psi$ as you pointed out. – Alex Provost Jul 08 '17 at 14:21
  • @AlexProvost Well, I am not sure there is no such way, all I was saying is that I do not see a way to do this... (Perhaps this line of inquiry is suitable for a separate question though). – Asaf Shachar Jul 08 '17 at 14:42

2 Answers2

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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.

Amitai Yuval
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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.

Thomas
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  • Thanks. Actually this was exactly my original argument! It's just that I introduced an inner product on $V$ to do this... but now (after seeing your solution) I think this is not necessary. – Asaf Shachar Jul 08 '17 at 07:07