I am attempting exercise #14 from Chapter 1 of Roman's Advanced Linear Algebra (p.57). Here is the statement of the problem:
Suppose $V$ is a finite-dimensional vector space over an infinite field $\mathbb{F}$. Suppose $V_1$, $V_2, \dotsc, V_n\leq V$ are subspaces of $V$ such that $\dim{V_1} = \dotsb = \dim{V_n}$. Show that there exists a subspace $S$ such that $V = V_i \oplus S$ for all $i$. (In other words, there exists a common complement for $V_i$.)
Now, if $\mathbb{F}$ is something familiar like $\mathbb{R}$ or $\mathbb{C}$ I can use measure-theoretic arguments to prove this. I might even be able to get away with it for any field of characteristic 0. But over something like $\mathbb{F}_2(t)$ or any other infinite field that is not obviously a measure space, that won't work. And that certainly is not the proof that Roman is looking for.
I think we want something like the proof he offers for Theorem 1.2 on p. 39, where he shows that a nontrivial vector space over an infinite field $\mathbb{F}$ is not a finite union of proper subspaces. Here he supposes $V = V_1 \cup \dotsb \cup V_n$ and for $w\in V_1 \setminus (V_2 \cup \dotsb \cup V_n)$ considers a set $A=\{rw + v \mid r \in \mathbb{F}\}$, and shows that each subspace must contain at most one vector from this set. Obviously this particular argument doesn't apply, but I think we want to make a similar argument.
Any ideas? Thanks
Best Answer
We can use downward induction on $\dim V_i$.
The case $\dim V_i=n$ is trivial (with $S=\{0\}$). Now suppose $\dim V_i<n$, then we have $\bigcup_iV_i\ne V$, so there is an $s\in V$ such that $s\notin \bigcup_iV_i$. Then apply the induction hypothesis on $V_i\oplus\Bbb F s$.