$V=W\oplus U=U\oplus X=W\oplus X$. Show there is a $2$ dim subspace $A\subseteq V$ s.t. each of $A\cap W, A\cap U$ and $A\cap X$ has dimension $1$.

Question

Question: Let $V$ be a vector space with subspaces $W,U, X$ such that $V=W\oplus U=U\oplus X=W\oplus X$. Prove there is a $2$ dimensional subspace $A\subseteq V$ such that each of $A\cap W, A\cap U$ and $A\cap X$ has dimension $1$.

Thoughts: I figured we could fix bases $\{w_1,\dots,w_k\}$, $\{u_1,\dots,u_n\}$, and $\{x_1,\dots,x_m\}$ for $W, U, X$, respectively such that $\dim W=k$, $\dim U=n$, and $\dim X=m$. Now, as $V=W\oplus U=U\oplus X=W\oplus X$, we have that $W\cap U=\{0\}$, $U\cap X=\{0\}, W\cap X=\{0\}$ (not sure if this helps, but since I want $A\cap\text{one of the subspaces}$ to have dimension $1$, I thought it might. I suppose Rank-Nullity may help here, but I wasn't quite seeing how. Maybe I should choose some two dimensional subspace $A$ of $V$, and then show the intersection stuff? I am just a bit stuck on how to do that. Any help is greatly appreciated! Thank you.

P.S. Could this be extended further to more than $3$ subspaces? Would the dimension of the intersection of $A$ and each subspace have to be adjusted?

Answer 1

First show that $W\neq 0$ by assuming $W=0$. This implies $X=V=U$ which would imply $V = 0$ (which presumably should be disallowed, otherwise we have a counterexample).

Let $w \in W$ be nonzero. Then $$w = w + 0 \in W \oplus U = U \oplus X,$$ so that there are $u \in U, x \in X$ such that $w = u + x$.

Also, $u \neq 0$, because if $u=0$, then we'd have $w = x \in X$ so $w \in X \cap W = \{0\}$, contradicting $w \neq 0$. For a similar reason, it follows that $x \neq 0$.

Now set $A = \operatorname{span}\{w, u\}$.

By construction, we immediately have $\operatorname{span}\{w\}, \operatorname{span}\{u\}, \operatorname{span}\{x\} \subseteq A$ so that $A\cap W, A\cap U, A\cap X$ all have dimension at least 1.

None has dimension $\geq 2$. For instance, if $A\cap W \subseteq A$ had dimension $\geq 2$, then we'd have $A\cap W = A$ and therefore $u \in A\cap U \subseteq W\cap U = \{0\},$ which contradicts $u \neq 0$.

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With regard to your postscript, it's not even clear how you intend to extend the problem to a larger number of subspaces. For instance, if you wanted to assume $$V = W_1\oplus W_2 = W_2\oplus W_3 = \ldots = W_{n-1} \oplus W_n = W_n \oplus W_1,$$ then the argument in my very first paragraph only works for $n$ odd. For $n$ even, you run into the problem that the sequence might just alternate $0, V, 0, V, \ldots$ so of course there's no one-dimensional subspace of $0$.

I've not thought about whether $n=5$ or higher odd values of $n$ might work.

Answer 2

Say $U\oplus W= V$ direct internal sum. A subspace $X$ such that $X\oplus W = V$ "is" the graph of a linear function $\phi \colon U\to W$. If we moreover want $X\oplus U = X$ then $\phi$ has to be bijective.

So now we have $U\oplus W = V$, $\phi\colon U\to W$ linear isomorphism and $X = \{u \oplus \phi(u)\ | u \in U\}$.

To get your subspace $A$: take $u\ne 0$ in $U$ and consider $A= \langle u, \phi(u)\rangle$. We have $A\cap U = \langle u\rangle$, $A\cap W = \langle \phi(u)\rangle$, $A \cap X = \langle u + \phi(u) \rangle$.