Show that $F_2^4$ is a union of three proper subspaces I'm just a bit confused about getting my head around this. I have seen proofs that say that a union of subspaces is only a subspace iff at least one subspace contains all the others, so, I'm not sure how this union of proper subspaces would work. Would somebody be able to show me how to do this?
There is also a follow-on question, show that there are 35 2-dimensional subspace of $F^4_2$ but I believe I can do this. We know that any 2-dimensional subspace is isomorphic to $F^2_2$, which itself has 6 choices of basis elements. Now, there are $(2^4 - 1)(2^4 - 2)$ total choices of two linearly independent vectors in $F^4_2$ and hence if we divide this total number of L.I. vectors by the result of 6 bases per subspace, we get the result of 35. Is this a valid method?
(NOTE: $F_2$ being the field {0,1})
Thanks, Helen.
 A: It is easily verified by listing elements that $F_2^4=V_1\cup V_2\cup V_3$ as follows:
\begin{align*}
V_1&=\langle(1,0,0,0),(0,1,0,0),(0,0,1,0)\rangle \\
V_2&=\langle(0,0,0,1),(0,0,1,0),(0,1,0,0)\rangle \\
V_3&=\langle(1,0,0,1),(1,1,0,1),(1,0,1,1)\rangle
\end{align*}
Maybe some insight into my approach will be helpful.  First I made the rather obvious choice for $V_1$.  From this I can obtain all vectors with last component $0$.  Next I made certain that not all vectors in $V_2$ have last component $0$.  Computing $V_1\cup V_2$, there are only three vectors not yet covered, which I chose as my basis for $V_3$.
A: Here's a picture of something similar happening in $F_2^3$, perhaps it will give you an idea about what can be done in general.

In the picture, we could think of our subspaces as the spans of $\{e_1, e_3\}, \{e_2, e_3\},$ and $\{e_1 + e_2, e_3\}$, to obtain the blue, green, and purple subspaces respectively.
A: The set where two coordinates are equal is a subspace, and two of the first three coordinates must be equal.
The proofs you refer to apply only to infinite fields.  Any finite vector space is clearly equal to the union of its dimension 1 subspaces.
