# Question about the intersection of a subspace of $V$ (which has basis $\mathcal{B}$) and a the spanning set of subset of its basis.

Let $$V$$ be a vector space with basis $$\mathcal{B}$$, $$\mathcal{A} \subseteq \mathcal{B}$$ and $$S$$ a subspace with basis $$\mathcal{C}$$. I think (am I'm trying to prove) that $$$$S \cap \operatorname{span} (\mathcal{A}) = \operatorname{span} (\mathcal{C}) \cap \operatorname{span} (\mathcal{A}) = \operatorname{span} (\mathcal{C} \cap \operatorname{span} (\mathcal{A}))$$$$ I think so since if we take all vectors as combinations from $$\mathcal{B}$$, we see that every $$c_i \in \mathcal{C}$$ is $$\sum a_i b_i$$, $$b_i \in \mathcal{B}$$. The consequence of this is that some vectors in the basis of $$S$$ might be linear combinations of the subset $$\mathcal{A}$$, if it contains enough vectors to generate this $$c_i$$. So, if $$c_i \in \operatorname{span} (\mathcal{A})$$, we have that the direct sum defining the intersection subspace will contain the span of $$c_i$$, since both subspaces will be able to generate this vector $$c_i$$ and its combinations by spanning. If some $$c_i$$ contains in its combinations an element of $$\mathcal{B}$$ that is not in $$\mathcal{A}$$, we know that the span of $$\mathcal{A}$$ won't be able to generate the $$c_i$$ and consequently none of its combinations. Hence, every basis for $$S$$ which is a linear combination of elements of $$\mathcal{A}$$ has its spanning contained in the spanning of $$\mathcal{A}$$, but any $$c_i$$ which cannot be generated by $$\operatorname{span} (\mathcal{A})$$ won't have its span contained in the span of $$\operatorname{span} (\mathcal{A})$$.

Sorry for the redundancy in my explanation.

So, are my assumptions correct?

• How are you defining the union of two subspaces? Do you mean the sum instead? (i.e.., $S_1 + S_2 = \{v+w : v \in S_1, w \in S_2\}$ rather than $S_1 \cup S_2$?) In particular, the third term in your equalities is a subspace, but the first two terms are not. – angryavian Dec 27 '19 at 21:03
• @angryavian thank you! I just realized I made a terrible mistake! Instead of intersections I put unions. – Gabriel C. Barbosa Dec 27 '19 at 21:06
• Have you looked at simple examples, like 2 planes in a 3d space ? – Carot Dec 27 '19 at 21:14
• @Carot If we take $V$ to be $\mathbb{R}^3$, and take $\mathcal{A} = \{ e_1, e_3 \}$ and $S$ to be $(1x, 1y, 0)$ spanned by $\{ e_1 + e_2 \}$. Then $S \cap \operatorname{span} (\mathcal{A}) = \{ 0 \} = \operatorname{span} (\mathcal{C} \cap \operatorname{span} ( \mathcal{A})) = \operatorname{span} (\varnothing)$. Also, for $\mathcal{A} = \{ e1, e2 \}$, we have by the intersection $\operatorname{span}(e1 + e2) = \{ (1x, 1y, 0) \} = S$, so my assumptions hold for y cases. – Gabriel C. Barbosa Dec 27 '19 at 21:34
• I cannot edit my previous comment but I mean my assumptions hold for "both examples" (I'm on mobile so sorry for the typos). – Gabriel C. Barbosa Dec 27 '19 at 21:45

For the second equality, let $$U := \text{span}(\mathcal{A})$$ so that we seek to verify $$\text{span}(\mathcal{C}) \cap U \overset{?}{=} \text{span}(\mathcal{C} \cap U)$$. It is not necessary to think about the basis $$\mathcal{A}$$.
It is also useful to recall the following characterization of span: for a set of vectors $$\mathcal{B}$$, its span $$\text{span}(\mathcal{B})$$ is the intersection of all subspaces containing those vectors.
The $$\supseteq$$ direction is straightforward: the left-hand side $$\text{span}(\mathcal{C}) \cap U$$ is a subspace that contains elements of $$\mathcal{C} \cap U$$, so in particular it contains $$\text{span}(\mathcal{C} \cap U)$$.
However, the other direction $$\subseteq$$ may not hold. A counterexample is $$U = \text{span}((1,1))$$ and $$\mathcal{C} = \{(1,0), (0,1)\}$$. The left-hand side is $$U$$ while the right-hand side is $$\{0\}$$.