Disjoint Union of Subsets and Direct Sum of Subspaces (Proof)

Let $B$ be a basis of the vector space $V$. Suppose $$B = B_1 \cup \cdots \cup B_n$$ for some subsets $B_{i}$ of $B$. Let $W_i = \mathrm{Span} (B_i)$.

Show that the following are equivalent:

• the $B_i$'s are pairwise disjoint, and
• $V = W_1 \oplus \cdots \oplus W_n$.

While I understand the intuition behind why the disjoint union of subsets and direct sum of subspaces are analogous, how should one prove this? How can I show that the intersection between the $W_i$'s are $\{ \boldsymbol{0} \}$?

• You surely mean that the $B_{i}$ are pairwise disjoint. – Andreas Caranti Feb 4 '17 at 14:15
• @AndreasCaranti, of course, edited to clarify. – Breton Thomas Feb 4 '17 at 14:20
• Would you mind if I edit your post for further clarity? You can always rollback my edit if you don't like it. – Andreas Caranti Feb 4 '17 at 14:22
• @AndreasCaranti Sure. Cheers. – Breton Thomas Feb 4 '17 at 14:24

First of all, the key point will be that given the assumption on the $B_{i}$, then for each $i$ we have $$W_{i} \cap (W_{1} + \dots + W_{i-1} + W_{i+1} + \dots W_{n}) = \{ 0 \}.$$ (It is not enough to show that the $W_{i}$ intersect pairwise in zero.)
And this is easy: if $$w_{i} = w_{1} + \dots + w_{i-1} + w_{i+1} + \dots w_{n}$$ for some $w_{i} \in W_{i}$, then $$w_{1} + \dots + w_{i-1} - w_{i} + w_{i+1} + \dots w_{n} = 0,$$ and now use the fact that $w_{i}$ is a linear combination of the elements of $B_{i}$, and that the $B_{i}$ are disjoint, and their union is a linearly independent set.