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 $N(B)=1$.

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    $\begingroup$ Let me denote your $d\psi_{\operatorname{Id}}(B))$ as $B^{+\wedge k}$ for the sake of brevity. Your fears are realized: $\operatorname{rank} B$ does not determine $\operatorname{rank} B^{+\wedge k}$. Indeed, if we let $k = 2$, then the latter rank is $1$ for $B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, and is $0$ for $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$; but of course both of these matrices $B$ have rank $1$. $\endgroup$ – darij grinberg Feb 12 at 0:24
  • $\begingroup$ Thank you; I actually wanted to restrict the discussion to the case where $k < \dim V$, but your comment put me on the right track. $\endgroup$ – Asaf Shachar Feb 12 at 11:52

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).


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