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$.


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 $k<d$, 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 $V$.

  • $\begingroup$ 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... $\endgroup$ – Asaf Shachar Mar 3 at 7:56
  • $\begingroup$ math.stackexchange.com/questions/3133267/…). $\endgroup$ – Asaf Shachar Mar 3 at 7:56
  • $\begingroup$ @AsafShachar Thanks. That's indeed a gap. Please see my new edit for a fix. $\endgroup$ – user1551 Mar 3 at 14:24
  • $\begingroup$ 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. $\endgroup$ – Asaf Shachar Mar 3 at 15:17
  • $\begingroup$ In other words, I don't see directly why in general if $Bv,Bu$ are linearly dependent, then $\lambda(u)=\lambda(v)$. $\endgroup$ – Asaf Shachar Mar 3 at 15:21

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