# Is a subspace the direct sum of all its intersections with a partition of the basis?

I've been working on this Linear Algebra problem for a while: Let $F$ be a field, $V$ a vector space over $F$ with basis $\mathcal{B}=\{b_i\mid i\in I\}$. Let $S$ be a subspace of $V$, and let $\{B_1, \dotsc, B_k\}$ be a partition of $\mathcal{B}$. Suppose that $S\cap \langle B_i\rangle\neq \{0\}$ for all $i$. Is it true that $S=\bigoplus\limits_{i=1}^{k}(S\cap \langle B_i \rangle)$?

Haven't been able to get this one, thanks for your help.

jgens
Gold Member

Notice that $V = \bigoplus_{i=1}^k \langle B_i \rangle$, so $S = S \cap V = \bigoplus_{i=1}^k S \cap \langle B_i \rangle$.

Notice that $V = \bigoplus_{i=1}^k \langle B_i \rangle$, so $S = S \cap V = \bigoplus_{i=1}^k S \cap \langle B_i \rangle$.
Where have you used that $S\cap \langle B_i\rangle \neq \{0\}$ for all $i$? I can come up with the following counterexample if we do not assume this hypothesis:

In $\mathbb{R}^2$, the subspace $y=x$ is certainly not the direct sum of its intersections with $\langle e_1 \rangle$ and $\langle e_2 \rangle$ (both zero).