$V_1, V_2$ subspaces of $V$. Show that there is a basis $\mathcal B$ for V such that $\mathcal B \cap (V_1 \cup V_2)= \emptyset$

Question

Let $V$ be a vector space over a field $F$. Suppose $V_1,V_2$ are subspaces of $V$ that lies strictly between $\{ 0_V \}$ and $V$. Show that there is a basis $\mathcal B$ for $V$ such that $\mathcal B\cap ( V_1 \cup V_2)=\emptyset $.

My work so far: My guess is that $V \setminus (V_1 \cup V_2)$ will make a linearly independent set and therefore can be extended to a basis for $V$. I showed that this set is non empty. So we can consider $S:=V\setminus (V_1 \cup V_2)$ How do I show that $S$ is linearly independent? Or is that even the right way to proceed?.

Any hints, suggestions? Thanks for your time.

Answer 1

This result does not hold in general. If $F=\Bbb F_2$ and $\dim V=2$ (whence $\dim V_i=1$ for $i=1,2$) and $V_1\neq V_2$, then $V\setminus(V_1\cup V_2)$ is a singleton, so you cannot choose a basis from its element.

The example is somewhat marginal, and the result ought to hold for any field with at least three elements, but this counterexample shows one will need to use the fact of having three distinct elements in your fields in a proof of that restricted statement. For $F=\Bbb F_2$ you can get counterexamples in any dimension by taking $V_1,V_2$ to be distinct hyperplanes (subspace of dimension $\dim V-1$); reduction modulo $V_1\cap V_2$ proves this (it reduces the question to the initial counterexample, since a basis must project to a spanning set of the quotient).