show that a set of vectors which contains a set of linearly dependent vectors is dependent? I was instructed to show that a set of vectors which contains a set of linearly dependent vectors is dependent. But what does this even mean? And how may I prove this?
What does it mean for a set to be linearly dependent? 
 A: Let $\vec {v_1},\vec {v_2}$ be two vectors. $\vec {v_1},\vec {v_2}$ being linearly dependant means there are scalars $c_1,c_2 \neq 0$ so that $c_1 \vec {v_1}+c_2 \vec {v_2}=0$. If they were linearly independent, then,  $c_1 \vec {v_1}+c_2 \vec {v_2}=0 \rightarrow c_1,c_2 = 0$
What linear dependance basically means is that you can express one vector (or sum of vectors) as the linear combination of the other vectors in the set. Linear combination meaning their sum with scaling them. 
So lets there be $v_1, v_2, \cdots, v_n$ a set of vectors. We know that $v_1, v_2, \cdots, v_k$ is linearly depandents. That means there are scalars $c_1,c_2,\cdots,c_k \neq 0$ so that $c_!v_1+c_2v_2+ \cdots +c_kv_k=0$
A set of vectors is linearly independent if $c_1v_1+c_2v_2+\cdots+c_nv_n=0 \rightarrow c_1=c_2=\cdots=c_n=0$, for every set of scalars. 
But we know that for $c_1, c_2 \cdots ,c_k, c_{k+1} \cdots, c_n$, as  $c_1,c_2,\cdots,c_k \neq 0$ and $c_{k+1}=0, c_{k+1}=0, \cdots c_n=0$ we'll gets $c_1v_1+c_2v_2+\cdots+c_nv_n=0$ without 
$c_1=c_2=\cdots=c_n=0$ (all of them that is), so the set is linearly dependant.  
A: Let's take 3 vectors $v_1,v_2,v_3$
If $v_3$ can be made by a linear combination of other two vectors then we say that this set of vectors is linearly dependent.
For example 
Let's take a set of vectors
${(1,0,1),(1,1,0),(2,1,1)}$
You can see a linear combination of first two vectors can make the third vector
$$k_1(1,0,1)+k_2(1,1,0)=(2,1,1)$$
Where $k_1=1,k_2=1$
You can see any multiple of this vector can be made by some linear combination of these two vectors.This means that any vector made by (2,1,1) can also be made by other two vectors. So, we say that this set is a linearly dependent set. Here (2,1,1) is a redundancy.
But if we remove it then the set becomes {(1,0,1)(1,1,0)}.Now this is linearly independent because no multiple of (1,0,1) can make any multiple of (1,1,0).
To check whether a set contains any redundancy, you can use reduced row echelon form. If all the elements of a line in rref becomes zero then that means that vector is redundant.
