0
$\begingroup$

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?

$\endgroup$
5
  • 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
1
$\begingroup$

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 $\{0\}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.