# Connection between ranks of an endomorphism and its linear image on the exterior power

Let $$V$$ be an $$n$$-dimensional real vector space, and let $$1. Let $$\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).

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 $$N(B)=1$$.

• 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 $1$. – darij grinberg Feb 12 at 0:24
• 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. – 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).