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Let $V$ be a real $d$-dimensional vector space, and let $1<k<d$ be fixed. Let $v_i$ be a basis for $V$. Consider the induced basis for the $k$-th exterior power $\bigwedge^k V$, given by $v_{i_1} \wedge \dots \wedge v_{i_k}$.

Suppose we have another basis $w_i$ for $V$, such that every $v_{i_1} \wedge \dots \wedge v_{i_k}$ equals to some $\lambda_{i_1,\ldots,i_k}w_{j_1} \wedge \dots \wedge w_{j_k}$ where $\lambda_{i_1,\ldots,i_k} \in \mathbb R$ are non-zero scalars. Is it true that after a possible rearrangement of the $w_i$, we must have $\text{span}(v_i)=\text{span}(w_i)$?

i.e. does there exist a permutation $\sigma \in S_d$ such that $\text{span}(v_i)=\text{span}(w_{\sigma(i)})$?

This question can be reformulated as a combinatorial question as follows:

By the independence of the $v_{i_1} \wedge \dots \wedge v_{i_k}$, for each strictly increasing multi-index $I=(i_1,\ldots,i_k)$, there corresponds a unique multi-index $J=(j_1,\ldots,j_k)$ such that

$v^I:=v_{i_1} \wedge \dots \wedge v_{i_k}=\lambda_Iw_{j_1} \wedge \dots \wedge w_{j_k}=\lambda_Iw^J$ holds. In other words, there exist a permutation $\tau \in S_{\binom{d}{k}}$ such that for every multi-index $I$, $v^I=\lambda_I w^{\tau(I)}$.

The question is whether or not $\tau$ is induced by a permutation $\sigma \in S_d$ in the obvious way: $\tau((i_1,\ldots,i_k))=(\sigma(i_1),\ldots,\sigma(i_k))$.

Indeed, if there exist such a permutation $\sigma$, then by setting $u_i:=w_{\sigma(i)}$, we are in the following situation:

$v_{i_1} \wedge \dots \wedge v_{i_k}=\lambda_Iu_{i_1} \wedge \dots \wedge u_{i_k}$, which implies $$\text{span}(v_{i_1},\dots,v_{i_{k-1}},v_{i_k})=\text{span}(u_{i_1},\dots,u_{i_{k-1}},u_{i_k}). \tag{1}$$ By switching between $i_k$ and $i_{k+1}$ in $(1)$,we obtain $$\text{span}(v_{i_1},\dots,v_{i_{k-1}},v_{i_{k+1}})=\text{span}(u_{i_1},\dots,u_{i_{k-1}},u_{i_{k+1}}). \tag{2}$$

By intersecting (1) and (2), we deduce that

$$\text{span}(v_{i_1},\dots,v_{i_{k-1}})=\text{span}(u_{i_1},\dots,u_{i_{k-1}}). \tag{3}$$

In the passage from $(1)$ to $(3)$ we have "removed" the last vectors $v_{i_k},u_{i_k}$. Continuing in this way, we can remove all vectors, until we get to $\text{span}(v_{i_1})=\text{span}(u_{i_1})$ as required.

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  • $\begingroup$ for which $k$ do you want your assumption to hold? since if you want it for all, everything collapses, and if you only want it for some, i think it is becoming far to weak (as it always holds for $k=n$) $\endgroup$ – Enkidu Mar 5 at 11:55
  • $\begingroup$ I want the assumption to hold for one specific fixed value of $k$, which is strictly between $1$ and $d=\dim V$. $\endgroup$ – Asaf Shachar Mar 5 at 13:39
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Yes, you can see it from basic geometry and combinatorics. Given a subset $I \subseteq [d]$, set $P_I = \operatorname{span} \{ v_i \}_{i \in I}$. Note that $P_I \neq P_J$ if $I \neq J$ and we have the nice property that $P_{I \cap J} = P_I \cap P_J$. Now, let's fix $1 \leq k < d$ and assume we are given the collections of subspaces $\{ P_I \}_{|I| = k}$ as a set (that is, without a specific order). Can we reconstruct the vectors $v_i$ up to permutation and scaling? Let's choose $n - 1 \choose k - 1$ subspaces which correspond to some collection of subsets $\mathcal{F}$ of size $k$ of $[n]$ and intersect them to get

$$ \bigcap_{I \in \mathcal{F}} P_I = P_{\bigcap_{I \in \mathcal{F}} I} $$.

We have only two options:

  1. The intesection $\bigcap_{I \in \mathcal{F}} I = \emptyset$ and then $\bigcap_{I \in \mathcal{F}} P_I = \{ 0 \}.$
  2. The intersection $\bigcap_{I \in \mathcal{F}} I = \{ i \}$ for some $1 \leq i \leq d$ and then $\bigcap_{I \in \mathcal{F}} P_I = P_{\{ i \}} = \operatorname{span} \{ v_i \}$.

The intersection $\bigcap_{I \in \mathcal{F}} I$ cannot contain two or more elements because the number of subsets which contain two specific elements is ${n - 2 \choose k - 2} < {n - 1 \choose k - 1}$. In addition, by running over all choices of subsets $\mathcal{F}$ of size ${n - 1 \choose k - 1}$ we will get all the subspaces $P_{\{ i \}}$ for $1 \leq i \leq d$.

Returning to your question, let's set $Q_I = \operatorname{span} \{ w_i \}_{i \in I}$. According to your assumption, the collections $\{ Q_I \}_{|I| = k}$ and $\{ P_I \}_{|I| = k}$ are identical and then by what I wrote above we must have that the collections $\{ Q_I \}_{|I| = 1}$ and $\{ P_I \}_{|I| = 1}$ are identical which implies that the $v_i$'s and the $w_i$'s are identical up to permutation and scaling.

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The answer is yes. Since there is no harm in doing so, we now work over an algebraically closed ground field (needed for the connectednesss argument).

Let $T$ be the torus in GL(V) with respect to the basis $\{v_i\}$ (so concretely $t\in T$ if $tv_i=t_iv_i$ for some nonzero scalars $t_i$).

Let $g\in GL(V)$ be the change of basis matrix $gv_i=w_i$.

Let $\phi:GL(V)\to GL(\wedge^kV)$ be the canonical homomorphism.

Suppose $t\in T$. Then $\phi(t)$ is diagonalisable with the $v_{i_1}\wedge\cdots\wedge v_{i_k}$ an eigenbasis. Also $\phi(gtg^{-1})$ is diagonalisable with the $w_{i_1}\wedge\cdots\wedge w_{i_k}$ an eigenbasis.

These eigenbases agree up to scalar. Therefore, for any $t,t'\in T$, the elements $\phi(t)$ and $\phi(gt'g^{-1})$ commute.

Now consider the map $f:T\times T\to GL(V)$ given by $f(t,t')=[t,gt'g^{-1}]$ (group commutator).

We have just demonstrated that the image lies in $\ker\phi$. Since $\ker\phi$ is discrete and $T\times T$ is connected, $f$ is constant. Since $f(1,1)=1$, that constant is 1.

Therefore for any $t\in T$, $gtg^{-1}$ lies in the centraliser of $T$. Since $T$ is its own centraliser, $gtg^{-1}\in T$. Therefore g lies in the normaliser of $T$, thus a product of an element of $T$ and a permutation matrix, as required.

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  • $\begingroup$ Thank you. This seems to be an interesting approach. However, there are a few things which I do not understand (assuming that by $\phi(A)$ you refer to the exterior power of $A$, i.e. $\phi(A)=\bigwedge^K A$) : (1) I don't see why the image of your map $f$ is contained inside $\ker \phi$. It seems to me that you are using something like $\phi([A,B])=[\phi(A),\phi(B)]$, so if $\phi(A),\phi(B)$ commute, then we have $[A,B] \in \ker \phi$. I don't think the equality $\phi([A,B])=[\phi(A),\phi(B)]$ holds in general, since $\phi$ is not additive, only multiplicative. $\endgroup$ – Asaf Shachar Mar 6 at 7:31
  • $\begingroup$ (2) I am pretty certain that $\ker \phi=0$...(when you restrict the domain the be $GL$), and I guess you meant to write $f(Id,Id)=0$, right? $\endgroup$ – Asaf Shachar Mar 6 at 7:31
  • $\begingroup$ $\phi(g)(u_1\wedge \cdots\wedge u_k)=gu_1\wedge \cdots \wedge gu_k$ is the definition of $\phi$ $\endgroup$ – Peter McNamara Mar 6 at 8:23
  • $\begingroup$ ok, but then why does the image of $f$ lies inside $\ker \phi$? As I said, $\phi$ is not linear, so it does not "respect" (preserve) commutators. $\endgroup$ – Asaf Shachar Mar 6 at 8:25
  • $\begingroup$ I'm not sure I understand your confusion. phi is a group homomorphism. The commutator is the group commutator aba^{-1}b^{-1}. $\endgroup$ – Peter McNamara Mar 13 at 3:45

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