How to prove that set $\{v_1-v_n,v_2-v_n,...,v_{n-1}-v_n\}$ is linearly independent if $\{v_1,v_2,...,v_n\}$ is a basis in $\mathbb{R^n}, n > 1$? 
Prove that set $U=\{v_1-v_n,v_2-v_n,...,v_{n-1}-v_n\}$ is linearly independent if $\{v_1,v_2,...,v_n\}$ is a basis in $\mathbb{R^n}, n > 1$.

I think one way to prove this is to suppose that for $k_1...k_n \in \mathbb R$:
$$
k_1(v_1-v_n)+k_2(v_2-v_n)_+...+k_{n-1}(v_{n-1}-v_n)=0
$$
Then we can rearrange the terms as:
$$
k_1v_1+k_2v_2+...+k_{n-1}v_{n-1}+v_n\cdot \Bigl( -\sum_{i=1}^{n-1}k_i\Bigr)=0
$$
Since $\{v_1,v_2,...,v_n\}$ is a basis in $\mathbb{R^n}$ then $\{v_1,v_2,...,v_n\}$ are linearly independent therefore $k_1=..=k_i..=k_{n-1}=0$ so even $\Bigl( -\sum_{i=1}^{n-1}k_i\Bigr)=0$ so $U$ is linearly independent.
Is my handling of sigma correct with the indices?
Lastly, could I've just used an elementary operation $(v_i-v_n)+v_n$ for $1 \le i \le n-1$ on the set $U$ to get the same result?
 A: Your reasoning is correct.
Also, you can indeed use elementary linear operations to get the same result. The full argument could for instance be as follows:
Note that every element of $\{v_1, \dots, v_n\}$ is a linear combination of elements of $\{v_1 - v_n, \dots, v_{n-1} - v_n, v_n\}$.
Because of this and because these sets have the same number of elements and because $\{v_1, \dots, v_n\}$ is a basis, $\{v_1 - v_n, \dots, v_{n-1} - v_n, v_n\}$ is a basis as well. (Or, formulating in terms of elementary operations, you've subtracted the $n$th basis vector from the first $n-1$ basis vectors, resulting in a new basis).
In particular, $\{v_1 - v_n, \dots, v_{n-1} - v_n, v_n\}$ is linearly independent and so the subset $\{v_1 - v_n, \dots v_{n-1} - v_n\}$ is linearly independent as well.
A: Alternatively, consider the linear map $\varphi:\mathbb{R}^n\to\mathbb{R}$ sending $\sum_{i=1}^n\,a_iv_i\mapsto \sum_{i=1}^n\,a_i$.  The kernel $K$ of $\varphi$ is an $(n-1)$-dimensional subspace of $\mathbb{R}^n$, since $\varphi$ is clearly surjective.  This kernel can be easily seen to equal to the span of $U$ as, for every $\sum_{i=1}^n\,a_iv_i\in K$, we have
$$\sum_{i=1}^n\,a_iv_i=\sum_{i=1}^{n-1}\,a_i\left(v_i-v_n\right)\,.$$
Being a spanning subset of an $(n-1)$-dimensional vector space and having $n-1$ elements, $U$ must be linearly independent.
A: The matrix having as columns the coordinates of the vectors with respect to the given basis is (it has $n$ rows and $n-1$ columns)
$$
\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 \\
0 & 1 & 0 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 0 & 1 \\
-1 & -1 & -1 & \dots & -1 & -1
\end{bmatrix}
$$
An obvious sequence of elementary row operations brings the matrix in the form
$$
\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 \\
0 & 1 & 0 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 0 & 1 \\
0 & 0 & 0 & \dots & 0 & 0
\end{bmatrix}
$$
showing its rank is $n-1$ (full column rank). So the given vectors form a linearly independent set.
Your proof is good as well.

You may want to try and find conditions so that, given the linearly independent set $\{w_1,\dots,w_m\}$ and the vector $x$, also the set
$$
\{w_1+x,\dots,w_m+x\}
$$
is linearly independent. Hint: let $U$ be the subspace spanned by $\{w_1,\dots,w_m\}$ and $U'$ be a complement of $U$; then write $x=y+z$, with $y\in U$ and $z\in U'$.
