linear independence of $\{x_j-x_i\}_{j=1 ,\\j \ne i}^{k+1}$? Show that if the set $\{x_2-x_1,...,x_{k+1}-x_1\} \subset \mathbb{R}^n$ is linearly independent then the following set is also linearly independent:
$$\{x_1-x_i,...,x_{i-1}-x_i,x_{i+1}-x_i,...,x_{k+1}-x_i\}$$
 A: Hint:
For fixed $i$, set up
$$
\sum_{j=1\\j\neq i}^{k+1}\mu_j(x_j-x_i)=0.
$$
Then by adding and subtracting $x_1$ inside the parentheses, we get
\begin{align}
0&=\sum_{j=1\\j\neq i}^{k+1}\mu_j((x_j-x_1)-(x_i-x_1)) \\
&=\sum_{j=2\\j\neq i}^{k+1}\mu_j(x_j-x_1)-(x_i-x_1)\sum_{j=1\\j\neq i}^{k+1}\mu_j.
\end{align}
A: I think it’s more simple than that, at first you should now that if you have a linear independent set, then a same size combination of this set it’s also l.i, so you just have to use the linear combination above and that’s it. The proof of this statement it’s essentially by contradiction.
Let’s have $\{x_1,\dots,x_n\}$ an l.i. set, want to proof that this happens if and only if $\{x_1,\dots,x_i+kx_j,\dots,x_n\}$ it’s a l.i. set for each $i,j$. Let’s make left to right, if $\{x_1,\dots,x_i+kx_j,\dots,x_n\}$ isn’t l.i. then we would have that for some index (not necessarily $i$ or $j$) it’s a linear combination of the rest, then:

*

*If $x_l$, for $l\in\{1,\dots,n\}-\{i,j\}$ it’s a linear combination of the other elements of the set then the contradiction it’s obvious because $x_l\in\{x_1,\dots,x_n\}$.

*Other wise the argument it’s pretty much the same, if $x_i+kx_j$ it’s a linear combination of the other elements in $\{x_1,\dots,x_i+kx_j,\dots,x_n\}$ then $x_i\in \{x_1,\dots,x_n\}$ it’s a linear combinaton of elements in that set.

So it’s clear that you can make finite linear combinations of a l.i. set and you will get another l.i. set.
