# Linear algebra intersections of sets and spans

$\newcommand{\sp}{\operatorname{sp}}$

$A,B$ are finite sets in a vector space $V$ over $F$.

Prove or disprove the following:

1. $\sp A \cap \sp B = \sp(A\cap B)$

2. $B\cap \sp A = \emptyset \Rightarrow A\cap \sp B= \emptyset$

For 1. the intersection of all the vectors that span $A$ and $B$ is the span of the intersection of $A$ and $B$. It seems so trivial that I find it difficult to formally prove...

As for 2. I couldn't find a counterexample but I'm not really sure on how to start.

Any help would be appreciated.

• The first statement is false. Take two vectors $A$ that span $R^2$ and take a set with different elements $B$ that also span $R^2$. The intersection of $A,B$ will be empty. While the intersection of span of $A$ and span $B$ is still $R^2$. – Maesumi Dec 4 '13 at 15:05

Hint: For 1) consider $A=\{\vec{v}\}$ and $B=\{2\vec{v}\}$. For 2) consider $A=\{(1,1)\}$ and $B=\{(1,0),(0,1)\}$ in $\mathbb{R}^2$.
• About 2. if you use these values, isn't $B\bigcap spA = A\bigcap spB$ ? – GinKin Dec 4 '13 at 15:17
• @GinKin: No. Since $sp B = \mathbb{R}^2$ and $sp A = \{ (a,a):a\in\mathbb{R} \}$, we obtain $B\bigcap spA = \emptyset$ and $A\bigcap spB = A$. – user35603 Dec 4 '13 at 22:21
1. Is false. Take $A=\{v\}$, $B=\{2v\}$ for some $v\neq 0$.
2. Is false. Take $v,w \in V$ independent vectors. Then take $A=\{v+w\}$ and $B=\{v,w\}$.
• I don't really see how does the intersections look like or what they equal to. We never saw things like this in class: $\{v,w\}\bigcap sp\{v+w\}$ OR $\{v+w\}\bigcap sp\{v,w\}$ Can you show what one of them is equal to ? – GinKin Dec 4 '13 at 15:26
• Try to prove that $v$ is not in $sp{v+w}$ if $v$ and $w$ are independent. Suppose on the contrary that $v$ is a multiple of $v+w$... what you get? – Emanuele Paolini Dec 4 '13 at 15:34