Exercise 12.2.1

Proof.

(a)

If there is no $v\in W_1\setminus (W_2\cup \cdots \cup W_m)$, then $W_1\subset W_2\cup \cdots W_m$. Therefore $V=W_2\cup \cdots W_m$, so $V$ is the union of $m-1$ proper subspaces, in contradiction with the definition of $m$. Thus there is $v\in W_1\setminus (W_2\cup \cdots \cup W_m)$.

(b)

There is $w\in V\setminus W_1$, since $W_1$ is a proper subspace. Since $v\in W_1\subset V$, and $w\in V$, $\mathit{\lambda v}+w\in V=W_1\cup \cdots \cup W_m$, for every $\lambda \in F$.

Let ${\mu }_1,\dots ,{\mu }_{m+1}$ be $m+1$ distinct elements of $F$. Since $F$ is infinite, it is possible to find such elements. For $i=1,\dots ,m+1$, ${\mu }_{i}v+w\in W_1\cup \cdots \cup W_m$.

Since there are more ${\mu }_i$ than subspaces $W_j$, there exist two distinct values ${\mu }_j\ne {\mu }_k$ such that ${\mu }_{j}v+w,{\mu }_{k}v+w$ are in the same subspace $W_i$. If we write ${\lambda }_1={\mu }_i,{\lambda }_2={\mu }_j$, then

$${\lambda }_1\ne {\lambda }_2,{\lambda }_{1}v+w=r\in W_i,{\lambda }_{2}v+w=s\in W_i\text{ for some }i,1\le i\le m.$$

(c)

Note that $W_i\ne W_1$, otherwise $w=r-{\lambda }_{1}v\in W_1$, and this is a contradiction with the definition of $w$. Therefore $$r-s=({\lambda }_1-{\lambda }_2)v\in W_i\text{ for some }i=2,\dots ,m.$$

Since ${\lambda }_1-{\lambda }_2\ne 0$, $v\in W_2\cup \cdots \cup W_m$, which is contradictory with the choice of $v$.

Conclusion: a finite dimensional vector space over an infinite field cannot be the union of a finite number of proper subspaces.