Proving vector space

Question

A real vector space$V$ is the union of three subspaces $P$$,$$Q$$,$$R$ of $V$.Prove that one of $P$$,$$Q$$,$$R$ is $V$. Please help . Give me some hints.

Answer 1

Let $P_1,\dots,P_n$ be proper subspaces of a real vector space $V$, then $\cup P_i \neq V$

Proof:

We prove by induction on $n$. The case $n=1$ is trivial. Assume the result is true for $n-1$, that means the union of any $n-1$ of $P_1,P_2,\dots, P_n$ cannot be $V$.

To prove the result for $n$, assume on the contrary that $V=\cup_1^n P_i$. Now I claim that we ihave $V=\cup_2^n P_i$, which contradicts the induction hypothesis.

In fact, we can pick $w\not \in V_1$, then for any $v\in V_1$, $v+cw\not \in V_1$ for any $c\neq 0$, but then $v+cw\in V_i$ for some $i\neq 1$. Since there are infinitely many $c$, some $V_i$($i\neq 1$) would contain some $v+c_1w$ and $v+c_2w$ ($c_1,c_2$ distinct non zero) and hence contain $v$. This proves that $V_1 \subset \cup_2^n V_i$

Answer 2

Let us generalize:

• We consider more than three subspaces
• We increase the subspaces until they are almost all of $V$
• We allow affine subspaces (i.e., translates of vecor subspaces)

While this makes the claim much stronger, it allows a simple proof:

Theorem. Let $V$ be a vector space over an infinite field $F$. For $1\le i\le n$, let $U_i\subset V$ be a 1-codimensional subspace and $v_i\in V$. Then $$ \bigcup_{i=1}^n(v_i+U_i)\ne V.$$

Proof. Recall that the intersection of two distinct 1-codimensional affine subspaces is either empty or 2-codimensional (i.e., 1-codimensional in each of the given subspaces).

Now we can prove the claim by (strong) induction on $n$, i.e., we assume that the claim is true if $n$ is replaces by any smaller integer and shall prove that it also holds for $n$:

If $n=0$, $\emptyset\ne V$ and we are done. So assume $n>0$. Pick $w\notin U_n$. Then for $a,b\in F$ with $a\ne b$, the affine subspaces $aw+U_n$ and $bw+U_n$ are disjoint because $(a-b)w\notin U_n$. Hence for at most $n$ out of infinitely many possible choices of $a$, it may happen that $aw+U_n=v_i+U_i$ for some $i$. So we can pick $a\in F$ such that this does not happen. Then $(v_i-aw+U_i)\cap U_n$ is either empty or 1-codimensional in $U_n$. Thus if we drop the empty terms (in particular, the one for $i=k$) from $$ \bigcup_{i=1}^n\bigl((v_i-aw+U_i)\cap U_n\bigr),$$ we obtain a union of less than $n$ 1-codimensional affine subspaces of $U_n$. By induction hypothesis, this union is $\ne U_n$ so that there exists $v\in U_n$ such that $v\notin v_i-aw+U_i$ for all $i$. Then $v+aw\notin v_i+U_i$ for all $i$. $\square$

---

Remark: The proof above would also work for a finite field $F$, as long as $n<|F|$.