Linear Dependence of n vectors We have vectors $$ v_{0}, v_{1}, ... v_{n}$$ in $$\ R^{m}$$ which are said to be affinely independent if with scalars $$c_{0},c_{1}...c_{n}$$ we have, $$\sum_{i=0}^{n}c_{i}v_{i}=0$$ and $$\sum_{i=0}^{n}c_{i}=0 \implies c_{i} = 0 \forall i = 0,1,...,n$$
If such is a case then which of the following is an implication
1) $$v_{0}, v_{1}, ... v_{n}$$ are linearly independent.
2) $$v_{1}-v_{0}, v_{2}-v_{0}, ..., v_{n}-v_{0}$$ are linearly independent.
3) $$n\leq m$$
The answer is given as , Only 2 and 3 are true.
Now, my doubt is, why doesn't statement 1 hold? Is it because if m = 2 (say) and the $$v_{i}$$ vectors have dimension 3 then they may not be linearly independent.
Also, does that imply linear dependence because if that happens then statement 2 is true
And finally, statement 3, how can we just say that $$n\leq m$$ just like that?
Please suggest a way. Thank you.
 A: To see why statement 1) is not true, you can consider $m=n=2$ and 
$$
v_0=\begin{pmatrix}1\\1\end{pmatrix},~v_1=\begin{pmatrix}1\\0\end{pmatrix},~v_2=\begin{pmatrix}2\\0\end{pmatrix}.
$$
From $c_0v_0+c_1v_1+c_2v_2=0$ we get
$$
c_1+2c_2=0\\c_0=0.
$$
From $c_0+c_1+c_2=0$ we conclude $c_1=-(c_0+c_2)=-c_2$ since $c_0=0$. This yields 
$$
c_1+2c_2=0\Leftrightarrow -c_2+2c_2=0\Leftrightarrow c_2=0.
$$
Finally we get $c_1=0$.
Therefore $v_0,v_1,v_2$ are affinely linear independent, but not lineare independent since $v_2=2v_1$.
For statement 2) you have to consider
\begin{align}
\sum_{i=0}^nc_i=0~\wedge~\sum_{i=0}^nc_iv_i=0
&\Leftrightarrow
c_0=-\sum_{i=1}^nc_i~\wedge~c_0v_0+\sum_{i=1}^nc_iv_i=0
\\&\Leftrightarrow c_0=-\sum_{i=1}^nc_i~\wedge~-\sum_{i=1}^nc_iv_0+\sum_{i=1}^nc_iv_i=0
\\&\Leftrightarrow c_0=-\sum_{i=1}^nc_i~\wedge~\sum_{i=1}^nc_i(v_i-v_0)=0.
\end{align}
Since we have
$$
\sum_{i=0}^nc_i=0~\wedge~\sum_{i=0}^nc_iv_i=0 \Rightarrow c_0=\ldots=c_n=0,
$$
we get
$$
\sum_{i=1}^nc_i(v_i-v_0)=0 \Rightarrow c_1=\ldots=c_n=0.
$$
Therefore $v_1-v_0,\ldots,v_n-v_0$ are linear independent.
The third statement follows directly from the second. If $n>m$ then $v_1-v_0,\ldots,v_n-v_0$ can't be lineare indepentent in $\mathbb{R}^m$, which contradicts statement 2.
