2

Let $V$ be a real $d$-dimensional vector space, and let $1\le k \le d-1$ be a fixed integer. Let $v \in V$ be non-zero, and denote by $X(v)$ the collection of all $k$-dimensional subspaces of $V$ that contain $v$. Observe that $\text{span}\{ v\}=\bigcap_{W\in X(v)}W.$

Here is my question:

Let $B:V \to V$ be a linear map. Is it true that $\bigcap_{W\in X(v)}BW \subseteq B(\text{span}\{ v\})\,$ ?

Since, the reverse inclusion clearly holds, this is equivalent to asking whether or not $\bigcap_{W\in X(v)}BW= B(\text{span}\{ v\})=B(\bigcap_{W\in X(v)}W)$.

If $B$ is invertible, then it commutes with intersections, so the answer is positive. What happens if $B$ is not invertible?

A positive answer to this question would establish a slick proof for this related problem.

Asaf Shachar
  • 25,967

1 Answers1

1

The answer is yes. Let $w\in V\setminus B($span$\{v\})$; we need to find $W\in X(v)$ so that $w\notin B(W)$. Let $C\colon V\to\Bbb R$ be a linear map with $C(w)=1$ and $C(B(v))=0$. ($C$ can be constructed, for example, by extending $\{B(v),w\}$ to a basis of $V$ and defining $C$ on each basis element. Here we use the fact that $w\notin B($span$\{v\})$.)

Then the kernel of $C\circ B\colon V\to\Bbb R$ has dimension at least $d-1$ and contains $v$. Let $W$ be a $k$-dimensional subspace of the kernel of $C\circ B$ that contains $v$, so that $W\in X(v)$. If $w\in B(W)$, then $1=C(w)\in C\circ B(W)=\{0\}$, a contradiction; therefore $w\notin B(W)$ as desired.

(The proof holds for vector spaces over any field.)

Greg Martin
  • 92,241
  • 1
    Thanks, this seems nice. I think you might intended to write $w \in V\setminus \text{span} Bv$ instead of $w \in V\setminus \text{span} v$, right? and similarly, $C(Bv)=0$ instead of $C(v)=0$. (This confused me...). – Asaf Shachar Mar 03 '19 at 09:41
  • @AsafShachar Thanks for spotting the errors! – Greg Martin Mar 03 '19 at 19:35