## 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 $$k. Thus, by lemma 1, $$\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 $$v_1 \in \ker A$$.