For linear subspaces of equal dimension, there exists a common complement

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

• A measure-theoretic argument? Start doing it when the $V_j$'s are lines in $\mathbb{R}^2$. You'll see that there is no need to invoke any measure. – Julien May 5 '13 at 1:46

2 Answers

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 $V_i\oplus\Bbb F s$.

• How do we know $\bigcup_i V_i \neq V$? I imagine this is where the hypothesis the field is infinite comes in...? – doodle Sep 16 '17 at 0:11
• Yes, exactly. This fact is included as known in OP. – Berci Sep 16 '17 at 0:27

Let $V$ be a finite dimensional vector space and let $U$ be a subspace of $V$.

If $C$ is a complement of $U$ in $V$, then there is a bijection $\phi$ from $\hom(C,U)$ to the set $\mathcal C(U)$ of all complements of $U$ in $V$: if $f\in\hom(C,U)$, then the subspace $\phi(f)$ is $\{c+f(c):c\in C\}$. You should be able to check that this indeed works without much problem.

Now suppose $U'$ is another subspace of $V$ of the same dimension as $U$. Show that the is a nn-zero poynomial function $\zeta:\hom(C,U)\to k$ such that $\zeta(f)\neq0$ iff $\phi(f)$ is a complement to both $U$ and $U'$.

From this, it follows that if $U_1$, $\dots$, $U_n$ are subspaces of $V$ of the same dimension and $C_1$ is a complement of $U_1$ in $V$, there is a non-zero polynomial function $\zeta:\hom(C,U)\to k$ such that whenever $f\in\hom(C,U)$ is not a zero of $\zeta$, then $\phi(f)$ is a complement to all of $U_1$, $\dots$, $U_n$.

Finally, on a positive dimensional vector space over an infinite field, a non-zero polynomial function does not vanish identically.