Is $S_1\cap S_2$ and $S_1\setminus S_2$ always linearly dependent if $S_1$ and $S_2$ are linearly dependent subsets of vector space $V$? Let $S_1$ and $S_2$ be linearly dependent subsets of vector space $V$, are $S_1\cap S_2$ and $S_1\setminus S_2$ always linearly dependent? 
The counterexample for the first one I can think of is $S_1=\{(1,0),(0,1), (1,1)\}$ and $S_2=\{(1,0),(0,1), (0,0)\}$.
The counterexample for the second one I can think of is $S_1=\{(1,0),(0,1), (1,1)\}$ and $S_2=\{(1,1)\}$.
But I am not sure if I'm correct?
 A: If $S_1$ and $S_2$ are dependent, neither $S_1 \cap S_2$ nor $S_1 \setminus S_2$ need to be dependent.
If $S_1 \cap S_2 = \emptyset$, then $S_1 \cap S_2$ is independent if you use the convention that $\emptyset$ is independent. 
If $S_1 \cap S_2$ contains a single vector $v \ne 0$, then $S_1 \cap S_2$ is independent. In particular, a set $(v)$ containing one vector is linearly independent if and only if $v \ne 0$.
If $S_1 \cap S_2$ contains a set of vectors $(v_1,\ldots,v_k)$, then $S_1 \cap S_2$ can be independent if $S_1 \cap S_2$ is a proper subset of $S_1$ and $S_2$.
Let $S_1=(v_1,\ldots,v_k)$ be a dependent set of vectors. Then $S_1$ can be made into a linearly independent set $S_1^\prime$ obtained by deleting all the vectors $v_j$ such that $v_j \in \text{span}(S_1) \setminus (v_j)$.
The set $S_1 \setminus S_2$ is independent if $S_1 \setminus S_2 = S_1^\prime$. In other words, $S_1 \setminus S_2$ is independent if $S_2$ is a dependent list containing all the vectors deleted from $S_1$ to obtain $S_1^\prime$.
A: You are correct.
Take any linearly independent set $A$, and let one of the elements be $a$. Then $A=(A\cup\{0\})\cap (A\cup 2a)$, and both of the sets we are intersecting are linearly dependent.
For the other example notice $A=(A\cup\{0\})\setminus\{0\}$

We assume the field is not of characteristic $2$ in the first example.
