Direct sums of subspaces Can someone check the correctness of my proof.
Statement.


*

*A single subspace $W_1$ is independent.

*Two subspaces $W_1,W_2$ are independent $\iff$ $W_1\cap W_2=\{0\}$


Two subspaces are said to be independent if $w_1+w_2=0, w_1\in W_1, w_2\in W_2$ implies that $w_1=0$ and $w_2=0$.
Proof.


*

*$0$ can be expressed uniquely once a basis for $W_1$ is chosen.

*Let $w\in W_1\cap W_2$ and let $B=(v_1,\dots,v_n),C=(w_1,\dots,w_m)$ be bases for $W_1,W_2$. Then $$w=\sum_i x_i v_i=\sum_j y_j w_j$$
$$0=w-w=\sum_i x_i v_i-\sum_j y_j w_j$$
As $B,C$ are bases, we have:
$$x_i=0\text{ }\forall i, y_j = 0\text{ } \forall j$$ 
Substituting back we get $w=0$ hence $W_1\cap W_2=\{0\}$


Conversely, suppose $W_1\cap W_2=\{0\}$. Consider the linear relation $$\sum_i x_i v_i + \sum_j y_j w_j$$
$$\sum_i x_i v_i = \sum_j (-y_j) w_j=v, \text{ for some } v$$
So $v=0\implies x_i = 0$ and $y_j=0$ for all $i,j$ because $B,C$ are bases. Therefor, $W_1,W_2$ are independent subspaces.

This argument extends to arbitrary collections of subspaces (by induction).
 A: Your proof for (2) is not entirely correct (but you're on the right path!). From
$$
0 = \sum_ix_iv_i - \sum_jy_jw_j,
$$ you cannot conclude that $x_i=y_j=0$ for all $i,j$ from the fact that $\{v_i\}$and $\{w_j\}$ are bases of $W_1$ and $W_2$ respectively. This assumes that $(v_1,\dots,v_n,w_1,\dots,w_n)$ is itself a basis... which you cannot assume at this point. Indeed, you're trying to show that $W_1\cap W_2=\{0\}$.
Instead:

Notice that $\sum_ix_iv_i\in W_1$ and $\sum_j(-y_j)w_j\in W_2$. As $W_1$ and $W_2$ are independent, this implies that
  $$
\sum_ix_iv_i = \sum_j(-y_j)w_j = 0
$$
  and so $w=0$. Therefore, $W_1\cap W_2=\{0\}$.

And conversely:

We consider the linear relation $v+w=0$ for some $v\in W_1$ and $w\in W_2$. Then $v=-w$. We can conclude that $v,w\in W_1\cap W_2$. By hypothesis $v=w=0$. By definition, $W_1$ and $W_2$ are independent.

Note that you began by writing a sum (expression) in $W_1+W_2$, but not a linear relation. Then how did you find that $\sum_ix_iv_i=\sum_j(-y_j)w_j$? And when you set $v=0$, you were essentially assuming that the summands were both $0$.
To see why your final remark on the generalization is false, you might try to find subspaces $W_1,W_2,W_3$ such that $\bigcap_iW_i=\{0\}$ but not all of
$$
W_1\cap W_2,\, W_1\cap W_3,\text{ and } W_2\cap W_3
$$
are trivial intersections.
