Let $V,W$ be $d$-dimensional real vector spaces of dimension $d$. Let $A \in \text{Hom}(V,W)$. Consider the induced map on the exterior algebras $\bigwedge ^kA:\bigwedge ^k V \to \bigwedge ^k W$. ($1 \le k \le d$).
It is easy to see* that $\text{rank} \bigwedge ^k A$ depends only on $\text{rank} \, A$. What is the function $F_k:\{0,\dots,d \} \to \{0,\dots,d \}$ such that $\text{rank} \bigwedge ^k A=F_k(\text{rank} \, A)?$
Partial results:
The cases $k=1,d$ are trivial.
$F_k(r) =0$ for $r < k$.
$F_k(k) =1$.
$F_k(s) \le \binom sk$ for $s \ge k$.
Indeed, suppose that $v_1,\dots,v_s$ are independent and that $A|_{\text{Span}\{ v_1,\dots,v_s\}}$ is injective. Complete $v_1,\dots,v_s$ to a basis of $V$. Then for any choice of $k$ basis elements, not all of them in $\{ v_1,\dots,v_s\}$ $\bigwedge ^k A$ gives zero when acting upon the corresponding "wedge-element".
- $F_k(s) = \binom sk$ for $s =k,d$. Note the case $s=d$ just means that if $A$ is invertible then $\bigwedge ^k A$ is invertible.
Summary:
It remains to compute $F_k(s)$ when $k<s<d$. I am not sure how to handle the possible linear dependencies of the images of $\bigwedge ^k A$ on different basis elements.
*Any two maps of the same rank are equivalent, and the exterior power is a functor. (So for invertible maps $(\bigwedge ^k A )^{-1}= \bigwedge ^k A^{-1}$ ).
Using Aaron's comment I am writing a proof:
Suppose $k \le s=\text{rank A} \le d$. Then, we can choose a bases $(w_1,\dots,v_d)$,$(v_1,\dots,w_d)$ for $V,W$, and define $A:V \to W$ by $Av_i=w_i$ for $i=1,\dots s$, $Av_j=0$ for $j=s+1,\dots,d$. Then the action of $\bigwedge^k A$ on the basis elements on $\bigwedge^k V$ are exactly $w_{i_1} \wedge \dots \wedge w_{i_k}$ where $1\le i_1 < i_2 < \dots < i_k \le s$ which are linearly independent.
This implies $\text{rank}(\bigwedge^k A)=\binom sk=\binom {\text{rank} A}{k}$.