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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} \} $?

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    You surely mean that the $B_{i}$ are pairwise disjoint. – 2017-02-04
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    @AndreasCaranti, of course, edited to clarify. – 2017-02-04
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    Would you mind if I edit your post for further clarity? You can always rollback my edit if you don't like it. – 2017-02-04
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    @AndreasCaranti Sure. Cheers. – 2017-02-04

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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.