# Does the action of a linear map on $k$-dimensional subspaces determine it up to scaling?

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$$.

Edit:

Here is a slick proof that the answer is positive in general:

Let $$v\in V$$ and let $$X(v)$$ be the collection of $$k$$-dimensional subspaces of $$V$$ that contain $$v$$. Then $$\text{span} \{v\}=\bigcap_{W\in X(v)}W,$$ so $$A(\text{span} \{v\})=A(\bigcap_{W\in X(v)}W) \subseteq \bigcap_{W\in X(v)}AW=\bigcap_{W\in X(v)}BW \subseteq B(\text{span} \{v\}),$$

where the last containment follows from this answer. This reduces the problem to the case where $$k=1$$.

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 $$v_1 \in \ker A$$.

The equality of kernels can be shown to always hold and then the argument via isomorphisms on quotient spaces works through:

Pick a basis $$v_1,\ldots, v_n$$ of $$\ker A\cap \ker B$$. Pick $$v^A_1,\ldots,v^A_m$$ such that together with the $$v_i$$, they form a basis of $$\ker A$$. Pick $$v^B_1,\ldots,v^B_m$$ such that together with the $$v_i$$, they form a basis of $$\ker B$$. (That the same $$m$$ occurs as for the $$v^A_i$$ follows from lemma 2). These $$n+2m$$ vectors are linearly independent: If $$\sum_i c_iv_i+\sum_ic^A_iv^A_i+\sum_ic^B_iv^B_i=0$$, then apply $$A$$ to find $$\sum_ic^B_iv^B_i\in \ker A\cap \ker B$$, hence all $$c_i^B=0$$. Likewise all $$c_i^A=0$$ and then all $$c_i=0$$. Hence we can pick $$u_1,\ldots, u_{d-n-2m}$$ such that $$\tag1v_1,\ldots, v_n, v^A_1,\ldots,v^A_m, v^B_1,\ldots,v^B_m, u_1,\ldots, u_{d-n-2m}$$ form a basis of $$V$$. Note that the $$Av_i^B$$ and $$Au_i$$ form a basis of $$V/\ker A$$, hence are linearly independent. Similarly, the $$Bv_i^A$$ and $$Bu_i$$ are linearly independent. Let $$W$$ be the subspace spanned by $$k$$ of the vectors in $$(1)$$. Then $$\dim A(W)$$ is the number of vectors picked from $$v_1^B\ldots v_m^B, u_1,\ldots, u_{d-n-2m}$$ and $$\dim B(W)$$ is the number of vectors picked from $$v_1^A\ldots v_m^A, u_1,\ldots, u_{d-n-2m}$$. By the given property, these numbers must be equal and hence also the same number of vectors were picked from the $$v_i^A$$ as from the $$v_i^B$$. If $$m>0$$ and $$k, it is clearly possible to violate this condition. We conclude $$m=0$$, i.e,

$$\ker A=\ker B.$$

Let $$D=\ker A=\ker B$$ and $$H=A(V)=B(V)$$. If $$D$$ has codimension $$\le 1$$, then $$\dim \operatorname{Hom}(V/D,H)\le 1$$ and so $$A,B$$ are linearly dependent. In all other cases, let $$k'=\max\{1,k-\dim D\}$$ and consider the isomorphisms $$\tilde A,\tilde B\colon V/D\to\operatorname{im}(A)$$ that $$A,B$$ induce. For any $$k'$$-dimensional subspace $$\tilde W$$ of $$V/D$$, we find a $$k$$-dimensional subspace $$W$$ of $$V$$ in the $$(k'+\dim D)$$-dimensional preimage and conclude $$\tilde A(\tilde W)=A(W)=B(W)\tilde B(\tilde W)$$. As $$1\le k'<\dim (V/D)$$, lemma 1 shows that $$\tilde A$$ and $$\tilde B$$ are linearly dependent, hence so are $$A$$ and $$B$$.

Since $$1\le k<\dim V$$, for every nonzero vector $$u$$, we have $$B\operatorname{span}(\{u\})=B\bigcap_{W\ni u,\,\dim W=k}W\subseteq\bigcap_{W\ni u,\,\dim W=k}BW=:\Delta.$$ The reverse inclusion is also true, for, if $$Bv\notin B\operatorname{span}(\{u\})$$, we may extend $$\{u,v\}$$ to a full basis $$\mathcal V\cup\mathcal K$$ of $$V$$ such that $$B\mathcal V$$ is a basis of $$BV$$ and $$\mathcal K$$ is a basis of $$\ker B$$. But if we pick $$k$$ vectors including $$u$$ but not $$v$$ from this full basis to form a set $$S$$, we would have $$Bv\notin B\operatorname{span}(S)\supseteq\Delta$$.

Hence $$B\operatorname{span}(\{u\})=\bigcap_{W\ni u,\,\dim W=k}BW$$ and the analogous holds for $$A$$. It follows from the assumption in your question that $$A\operatorname{span}(\{u\})=B\operatorname{span}(\{u\})$$ for every nonzero vector $$u$$. Consequently, $$\ker A=\ker B$$ and $$AV=BV$$.

Thus the restrictions $$f$$ and $$g$$ of $$A$$ and $$B$$ respectively on $$\operatorname{span}(\mathcal V)$$ are bijective linear maps between $$\operatorname{span}(\mathcal V)$$ and $$AV(=BV)$$. Also, $$f\left(\operatorname{span}(\{u\})\right)=g\left(\operatorname{span}(\{u\})\right)$$ for every nonzero vector $$u\in\operatorname{span}(\mathcal V)$$. But that means every nonzero vector $$u\in\operatorname{span}(\mathcal V)$$ is an eigenvector of $$g^{-1}\circ f$$. Therefore $$g^{-1}\circ f=\lambda\operatorname{Id}$$ for some scalar $$\lambda$$, meaning that $$A=\lambda B$$ on $$V$$.

• Thanks. I think there might be a problem with your approach: It is not clear to me that the assumption indeed implies that the images of $A$ and $B$ on one-dimensional subspaces are the same. The problem is that this argument seems to use the claim that $\bigcap_{W\in X(v)}BW \subseteq B(\text{span}\{ v\})$, (where $X(v)$ is the collection of all $k$-dimensional subspaces of $V$ that contain $v$).I am not sure that this claim is holds when $B$ is non-invertible. (Indeed, I came to wonder about this point even before I saw your answer; I have now asked a separate question about it here... – Asaf Shachar Mar 3 at 7:56
• – Asaf Shachar Mar 3 at 7:56
• @AsafShachar Thanks. That's indeed a gap. Please see my new edit for a fix. – user1551 Mar 3 at 14:24
• Thanks. I also think that the argument for the case where $k=1$ would be more clear if you will take a basis for $\ker A=\ker B$ and then complete it into a basis of $V$; Then on the "remaining part" of the basis, $A,B$ would be injective, which implies that $Bv,Bu$ are linearly independent whenever $u,v$ are linearly independent. – Asaf Shachar Mar 3 at 15:17
• In other words, I don't see directly why in general if $Bv,Bu$ are linearly dependent, then $\lambda(u)=\lambda(v)$. – Asaf Shachar Mar 3 at 15:21