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.

  • $\begingroup$ 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 $k=n$) $\endgroup$
    – Felix
    Commented Mar 5, 2019 at 11:55
  • $\begingroup$ I want the assumption to hold for one specific fixed value of $k$, which is strictly between $1$ and $d=\dim V$. $\endgroup$ Commented Mar 5, 2019 at 13:39

2 Answers 2


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.


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.

  • $\begingroup$ 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. $\endgroup$ Commented Mar 6, 2019 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? $\endgroup$ Commented Mar 6, 2019 at 7:31
  • $\begingroup$ $\phi(g)(u_1\wedge \cdots\wedge u_k)=gu_1\wedge \cdots \wedge gu_k$ is the definition of $\phi$ $\endgroup$ Commented Mar 6, 2019 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. $\endgroup$ Commented Mar 6, 2019 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}. $\endgroup$ Commented Mar 13, 2019 at 3:45

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