linear independence after subtraction of vector Given a set $\{v_0, v_1, v_2\}$ of linearly independent vectors, what happens if I subtract each vector of that set by, say, $v_0$?
Then I obtain the set $\{0, v_1-v_0, v_2-v_0\}$. However, since any family of vectors containing the zero vector is dependent, this would mean that our new set is now dependent, right? But that again would mean that the dimension of the first set was 3 and that of the second is 2. So this would mean that I can reduce the dimension of a set by subtracting vectors?
Now, if I look at $span\{v_0, v_1, v_2\}$ and $span\{v_0, v_1, v_2\} - v_0$. Is the latter set equal to $span\{0, v_1-v_0, v_2-v_0\}$ and does this set again have one dimension less than the span without subtracting a vector?
 A: Notice that $span\{v_0, v_1, v_2\}-v_0=span\{v_0, v_1, v_2\}$, since a generic element of the former is of the form $a_0v_0+a_1v_1+a_2v_2-v_0=(a_0-1)v_0+a_1v_1+a_2v_2$, and any linear combination of $v_0, v_1, v_2$ can be expressed in this way.
So $span\{v_0, v_1, v_2\}-v_0$ has dimension $3$, while $span\{0, v_1-v_0, v_2-v_0\}=span\{v_1-v_0, v_2-v_0\}$ has dimension $2$, hence they cannot be equal.
A: $span\{v_0,v_1,v_2\}$ is a vector space of dimension $3$ while $span\{v_0-v_0,v_1-v_0,v_2-v_0\}$ is always a vector space of dimension $3-1=2$ but $span\{v_0,v_1,v_2\}-v_0$ is not even a vector space, due to the lack of closure property with respect to the addition. For instance; $span\{(1,0),(0,1\}-(1,0)=\mathbb R^2-(1,0)$ is not a vector space as it misses the sum  $(1,1)+(0,-1)$.
EDIT: After one of the users pointed out that OP defined $A-v=\{a-v:a\in A\}$, here is the modified part of my original answer. The set $span\{v_0,v_1,v_2\}-v_0$ is essentially the same vector space as $span\{v_0,v_1,v_2\}$ due to the closure property between vector '$-v_0$' and an arbitrary vector of $span\{v_0,v_1,v_2\}$ which implies the equality of two sets.
