If the images of two linear maps coincide on $k$-dimensional subspaces, are they proportional?

Let $ V$ be a real $ d$ -dimensional vector space, and let $ 1 \le k \le d-1$ be a fixed integer. Let $ A,B \in \text{Hom}(V,V)$ , and suppose that $ AW=BW$ for every $ k$ -dimensional subspace $ W \le V$ . Is it true that $ A=\lambda B$ for some $ \lambda \in \mathbb R$ ? If not, can we characterize all such pairs $ A,B$ ?

Here are some partial results (proofs at the end):

First, the answer is clearly positive for $ k=1$ .

Lemma 1: If at least one of $ A$ and $ B$ is invertible, then the answer is positive.

Lemma 2: We always have $ \text{Image}(A)=\text{Image}(B)$ . In particular, $ \text{rank}(A)=\text{rank}(B)=r$ .

Lemma 3: If $ r \ge k$ or $ r \le d-k$ , then $ \ker(A)=\ker(B) $ .

In particular, the above lemmas imply that if $ r>k$ , then the answer is positive. Indeed, in that case, the kernels and images coincide, so we can consider the quotient operators: $ \tilde A,\tilde B:V/D \to H$ , where $ D$ is the kernel, and $ H$ is the image. Now $ \tilde A, \tilde B$ are invertible operators between $ r$ -dimensional spaces, and they satisfy the assumption for $ k<r$ . Thus, by lemma 1, $ \tilde A=\lambda \tilde B$ , which implies $ A=\lambda B$ .


Proof of Lemma 1:

Suppose that $ A$ is invertible. Then, we have $ SW=W$ , where $ S=A^{-1}B$ . Thus, every $ k$ -dimensional subspace is $ S$ -invariant, which implies $ S$ is a multiple of the identity.

Proof of Lemma 2: $ \text{Image}(A)=\text{Image}(B)$ .

Let $ x=Av_1 \in \text{Image}(A)$ ; complete $ v_1$ to a linearly independent set $ v_1,\dots,v_k$ . Then $ $ x \in A(\text{span}\{v_1,\dots,v_k\})=B(\text{span}\{v_1,\dots,v_k\})\subseteq \text{Image}(B),$ $

so $ \text{Image}(A) \subseteq \text{Image}(B)$ . The other direction follows by symmetry.

Proof of Lemma 3: If $ r \ge k$ or $ r \le d-k$ , then $ \ker(A)=\ker(B) $ .

First, suppose that $ r \ge k$ , and let $ v_1 \notin \ker A$ . Complete $ v_1$ into a linearly independent set $ v_1,\dots,v_k$ such that $ A(\text{span}\{v_1,\dots,v_k\})$ is $ k$ -dimensional. Then $ B(\text{span}\{v_1,\dots,v_k\})$ is $ k$ -dimensional, so $ Bv_1 \neq 0$ . This shows $ \ker(A)^c \subseteq \ker(B)^c$ , i.e. $ \ker(B)\subseteq \ker(A)$ . The other direction follows by symmetry.

Now, suppose that $ r \le d-k$ . Then, since the nullity is $ \ge k$ , every $ v_1 \in \ker B$ can be completed into a linearly independent set $ v_1,\dots,v_k$ , all in $ \ker B$ . This implies that $ A(\text{span}\{v_1,\dots,v_k\})=0$ , so $ v_1 \in \ker A$ .