Let $V$ be an $n$-dimensional real vector space, and let $1<k<n$. Let $\psi:\text{End}(V) \to \text{End}(\bigwedge^kV)$ be the exterior power map, $\psi(A)=\bigwedge^k A$.
For $B \in \text{End}(V)$, we denote $B^{+\wedge k}=d\psi_{\operatorname{Id}}
(B) \in \text{End}(\bigwedge^kV)$: $$ B^{+\wedge k}(v_1 \wedge \dots \wedge v_k)=\sum_{i=1}^k v_1 \wedge \dots \wedge v_{i-1} \wedge Bv_i \wedge v_{i+1}
\wedge \ldots \wedge v_k. $$
I know that $B=0 \iff B^{+\wedge k}=0$. (i.e. $\psi$ is an immersion, here we use the assumption $k<n$).
Moreover, $N(B):=\dim(\ker B)=r\ge k \Rightarrow N(B^{+\wedge k})\ge \binom{r}{k}$.
Indeed, choose a basis $v_1,\dots,v_r$ for $\ker B$; then every $v_{i_1} \wedge \dots \wedge v_{i_k} \in \ker(B^{+\wedge k})$, where $1 \le i_1 < \ldots < i_k \le r$, and these are linearly independent.
Question: Does the converse implication hold? Does $ N(B^{+\wedge k})\ge \binom{r}{k} \Rightarrow N(B)\ge r$? (for $r \ge k$)
Part of the problem is that I am not sure that $\ker(B^{+\wedge k})$ admits a basis consisting of decomposable elements.
Note that $\text{rank}(B)$ does not determine $\text{rank}(B^{+\wedge k})$. Furthermore, $B^{+\wedge k}$ can have full rank while $B$ does not.
Here are examples:
Example 1:
Let $n=3,k=2$, and let $v_1,v_2,v_3$ be a basis for $V$. Set $Bv_1=v_1,Bv_2=0,Bv_3=v_3$. Then, after identifying $\bigwedge^2 V$ with $\mathbb{R}^3$ via $$a(v_1 \wedge v_2)+b(v_1 \wedge v_3)+c(v_2 \wedge v_3) \cong \begin{bmatrix} a \\ b \\ c \end{bmatrix}, \, \, \text{we have} $$
$B^{+\wedge k}\left(\begin{bmatrix} a \\ b \\ c \end{bmatrix}\right)=\begin{bmatrix} a \\ 2b \\ c \end{bmatrix}$, so $B^{+\wedge k}$ is invertible.
Example 2:
Setting $Bv_1=0,Bv_2=v_1,Bv_3=v_3$, we get $B^{+\wedge k}\left(\begin{bmatrix} a \\ b \\ c \end{bmatrix}\right)=\begin{bmatrix} 0 \\ b+c \\ c \end{bmatrix}$, so $N(B^{+\wedge k})=1$. In both cases $N(B)=1$.
Best Answer
The answer is negative. Let $n=3,k=2$, and let $v_1,v_2,v_3$ be a basis for $V$. Set $Bv_1=v_2,Bv_2=v_1,Bv_3=v_3$. After identifying $\bigwedge^2 V$ with $\mathbb{R}^3$ via $$a(v_1 \wedge v_2)+b(v_1 \wedge v_3)+c(v_2 \wedge v_3) \cong \begin{bmatrix} a \\ b \\ c \end{bmatrix}, \, \, \text{we have} $$ $B^{+\wedge 2}\left(\begin{bmatrix} a \\ b \\ c \end{bmatrix}\right)=\begin{bmatrix} 0 \\ b+c \\ b+c \end{bmatrix}$, so $N(B^{+\wedge 2})=2\ge\binom{2}{2}$, but $B$ is invertible.
In particular, this shows that the invertibility of $B$ does not imply the invertibility of $B^{+\wedge k}$, and vice versa. (One direction was demonstrated in the question).