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 \begin{equation} S \cap \operatorname{span} (\mathcal{A}) = \operatorname{span} (\mathcal{C}) \cap \operatorname{span} (\mathcal{A}) = \operatorname{span} (\mathcal{C} \cap \operatorname{span} (\mathcal{A})) \end{equation} 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?

  • 1
    $\begingroup$ 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. $\endgroup$ – angryavian Dec 27 '19 at 21:03
  • $\begingroup$ @angryavian thank you! I just realized I made a terrible mistake! Instead of intersections I put unions. $\endgroup$ – Gabriel C. Barbosa Dec 27 '19 at 21:06
  • $\begingroup$ Have you looked at simple examples, like 2 planes in a 3d space ? $\endgroup$ – Carot Dec 27 '19 at 21:14
  • $\begingroup$ @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. $\endgroup$ – Gabriel C. Barbosa Dec 27 '19 at 21:34
  • $\begingroup$ I cannot edit my previous comment but I mean my assumptions hold for "both examples" (I'm on mobile so sorry for the typos). $\endgroup$ – Gabriel C. Barbosa Dec 27 '19 at 21:45

The first equality holds by definition.

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 $\{0\}$.


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