The wedge of two linear maps has any meaning?

Question

$\newcommand{\Hom}{\operatorname{Hom}}$

Let $V,W$ be $d$-dimensional real vector spaces, $A,B \in \Hom(V,W)$.

Let us try to define a map $A \wedge B:\Lambda_2(V) \to \Lambda_2(W)$, by setting $$ A \wedge B (v \wedge w)=Av \wedge Bw. \tag{1}$$

(This imitates the standard construction $A \wedge A (v \wedge w)=Av \wedge Aw$).

The problem with $(1)$ is that $A \wedge B$ is well-defined only when $$ Av \wedge Bw=-Aw \wedge Bv \tag{2}.$$

This condition is equivalent to

$$ Av \wedge Bv=0 \, \, \text{ for every} \, v \in V \tag{3},$$

which is equivalent to $$ \dim({\operatorname{span}\{Av,Bv\}}) \le 1 \, \, \text{ for every} \, v \in V . \tag{4}$$

Questions:

$(1)$ While Condition $(4)$ is very "geometric" (the images of every vector lies on a single line), it is not clear to me how to check it in practice (it's not enough to check on basis vectors).

Are there any "nice" necessary or sufficient conditions for it to hold? ($A =\lambda B$ suffices of course, but this is very trivial).

Edit: It turns out (see my answer below) that if $\max(\operatorname{rank}(B),\operatorname{rank}(A)) \ge 2$ then it is necessary that $A$ is multiple of $B$.

If at least one of the ranks is zero, we are done (one operator is zero, the other can be chosen arbitrarily). Otherwise, $\operatorname{rank}(B)=\operatorname{rank}(A)=1$. This is still open for now.

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$(2)$ In case where $A \wedge B$ is well-defined, does it have a "geometric meaning"? (The classic $A \wedge A$ represents the action of $A$ on $2$-dimensional parallelepipeds).

Answer 1

There's an alternative construction which always makes sense: map $v\wedge w$ to $$Av\wedge Bw+Bv\wedge Aw.$$ I suppose one gets that from "polarizing" the $\bigwedge^2A$ and $\bigwedge^2B$ where $\bigwedge^2A(v\wedge w)=Av\wedge Aw$: $$Av\wedge Bw+Bv\wedge Aw= \left(\bigwedge{}^2(A+B)-\bigwedge{}^2A-\bigwedge{}^2B\right)(v\wedge w).$$

Answer 2

This is a partial answer:

Proposition: Suppose that $\operatorname{rank}(B) \ge 2$ or $\operatorname{rank}(A) \ge 2$, and that $\dim({\operatorname{span}\{Av,Bv\}}) \le 1 \, \, \text{ for every} \, v \in V . $ Then $A = \lambda B$ for some $\lambda \in \mathbb{R}$.

Proof:

Since the condition is symmetric w.r.t $A,B$, we can assume W.L.O.G that $\operatorname{rank}(B) \ge 2$.

Define $D=\ker B$, and let $D^{\perp}$ be a direct summand of $D$ (we can put a metric on $V$, and take $D^{\perp}$ to be the orthogonal complement, but it does not matter).

Let $v\in D^{\perp}$ be a non-zero vector. By assumption, there is a non-trivial linear combination $\lambda Av +\tilde \lambda Bv=0$. Suppose $\lambda =0$. Then $Bv=0$ so $v=0$, a contradiction. Hence, we can divide by $\lambda $, and write $Av=cBv$ for some $c \in \mathbb{R}$ (depending on $v$).

Let $v_1,v_2 \in D^{\perp}$ be two independent vectors (here we use $\operatorname{rank}(B) \ge 2)$. Since $B|_{D^{\perp}}$ is injective, $Bv_1,Bv_2$ are independent.

By what we just proved there exist $\lambda_1,\lambda_2$ s.t

$$ Av_i=\lambda_iBv_i.$$ Then $w=A(v_1+v_2)=\lambda_1Bv_1+\lambda_2Bv_2$ and $\tilde w=B(v_1+v_2)=Bv_1+Bv_2$ are linearly dependent.

Hence $\lambda w +\tilde \lambda \tilde w=0$.

Note that $\tilde w \neq 0$ (since $B|_{D^{\perp}}$ is injective and $v_1,v_2 $ are independent).

So, $\,\lambda \neq 0$, thus $$w=c\tilde w \Rightarrow \lambda_1Bv_1+\lambda_2Bv_2=cBv_1+cBv_2. $$

Since $Bv_1,Bv_2$ are independent, this forces $\lambda_1=\lambda_2=c$.

This shows $A|_{D^{\perp}}=cB|_{D^{\perp}}$. By definition of $D$, $B|_D=0$. We show $A|_D=0$:

Let $v \in D$, and take some non-zero $w \in D^{\perp}$.

Then, $A(v+w)=Av+cBw,B(v+w)=Bw$ are dependent. So

$$ \lambda (Av+cBw )+ \tilde \lambda Bw=0.$$

As before $\lambda \neq 0$, so $Av = \tilde c Bw$ for some $\tilde c$.

But $w$ was an arbitrary non-zero vector in $D^{\perp}$.

Since $\operatorname{rank}(B)=\dim (D^{\perp}) \ge 2$, then we can take two independent $w_1,w_2 \in D^{\perp}$, and conclude $Av=0$ as required.