I'm trying to solve this problem: Suppose that $V$ is a vector space with basis $\mathcal{B}=\{b_i:i\in I\}$ and $S$ is a subspace of $V$. Let $\{B_1,\ldots,B_k\}$ be a partition of $\mathcal{B}$. Then is it true that $S=\bigoplus_{i=1}^{k}(S\cap \left<B_i\right>)$? What if $S\cap \left<B_i\right>\neq \{0\}$ for every $i$?
For the first part I saw that is false, because we can consider for example $V=\mathbb{R}^2$, $S=\left<(1,1)\right>$, $\mathcal{B}=\{(1,0),(0,1)\}$, $B_1=\{(1,0)\}$ and $B_2=\{(0,1)\}$. In this case we have that $S\cap \left<B_i\right>=\{0\}$ for $i=1,2$ and so $S\neq\bigoplus_{i=1}^{2}(S\cap \left<B_i\right>)$. But for the case that $S\cap \left<B_i\right>\neq \{0\}$ I don't know if this is true or not. I supposed that it was true and tried to prove it, but in some part I couldn't continue since some vectors that I obtained not necessarily are in $S$. And I haven't found a counterexample. Could you please give me some suggestions for this part? Thanks.
Consider $\mathbb{R}^4$ and $S=\langle(1,1,0,0),(0,0,1,0),(0,0,0,1)\rangle$.
Consider the following partiron of the canonical basis: $B_3=\{e_1, e_3\}$ and $B_4=\{e_2, e_4\}$
where $e_i$ has all the coordinates equal to zero except for a 1 in the position $i$.
Is easy to check that $S\cap \langle B_j\rangle=\langle e_j\rangle$, obtaining a counter example.
Moreover, I have to mention that if you impose $S\cap \langle B_i\rangle\neq \{0\}$ for all $i$, your claim is true in $\mathbb{R}^n$ if (and only if) $n \leq 3$ by some easy arguments of dimentions of the intersections.