# Let $V$ be a finite dimensional vector space over $R$, then is it possible to write $V$ as union of finitely many proper subspaces? [duplicate]

Let $$V$$ be a finite dimensional vector space over $$R$$, then is it possible to write $$V$$ as union of finitely many proper subspaces?

I am not sure but suppose if we have $$B$$ as an ordered basis for our vector space $$V$$, then we may break $$B$$ into $$B_1$$ and $$B_2$$ such that $$B1\cap B2=\phi$$ and $$B1\cup B2=B$$

So, isn't it true that $$V=L(B1) \cup L(B2)$$?

Where $$L(B1)$$ is linear span of $$B1$$

• It's true with $+$ instead of $\cup$ on the right hand side. Think about the cross in 2d. – Berci Sep 25 '19 at 22:27
• It is possible if and only if it is a space of dimension $\geq 2$ over a FINITE field. If the dimension is 1 or 0, then any subspace is improper. – N. S. Sep 25 '19 at 23:54
• This question has been asked, and answered, several times on this website. – Gerry Myerson Sep 26 '19 at 0:39

The Lebesgue measure we have $$mB = 0$$ for any proper subspace. Hence any countable or finite union of proper subspaces $$\cup_n B_n$$ has measure zero. Since $$m\mathbb{R}^n = \infty$$ we see that it cannot be written as a countable or finite union of proper subspaces.

There is a general argument: consider an infinite field $$K$$ and $$V$$ any $$K$$-vector space.

Then $$V$$ is not covered by finitely many proper subspaces $$W_i$$.

The proof is by induction on the number $$p$$ of subspaces. If $$p=1$$, there is nothing to prove. If the statement holds till rank $$p-1$$, and $$V$$ is covered by the proper subspaces $$W_1, \ldots W_p$$, there is some $$x \in W_1$$ that does not belong to any $$W_k$$, $$k\geq 2$$ (else the $$W_k$$, $$k \geq 2$$, cover $$V$$, impossible by induction hypothesis).

Of course, there is some $$y \in W_2$$ but not in $$W_1$$ (else, the induction hypothesis applies to $$W_1, W_3,\ldots, W_p$$). Consider $$\Delta=\{z_t=(1-t)x+ty,\,t \in K\}$$: it has infinitely many points so meets some $$W_t$$ at least twice. Since $$W_t$$ is a proper subspace, $$\Delta \subset W_t$$. So $$x \in W_t$$ so $$t=1$$. But also $$y \in \Delta$$ so $$y \in W_t$$, so $$t \neq 1$$. We have a contradiction.

No.

Take for instance $$\Bbb{R}^2$$

The only proper subspaces of this space,are the lines that pass through the origin and $$\{(0,0)\}$$