# For vectors $a_1 - a_n, \ldots, a_{n-1} - a_n$ linearly independent $\Leftrightarrow$ $a_2 - a_1, \ldots, a_n - a_1$ linearly independent

Let $$a_1, \ldots, a_n$$ for $$n \geq 2$$ vectors in an $$\mathbb{F}$$-vector space $$V$$.

Show the following:

$$a_1 - a_n, \ldots, a_{n-1} - a_n$$ linearly independent $$\Leftrightarrow$$ $$a_2 - a_1, \ldots, a_n - a_1$$ linearly independent

## My (flawed) idea

I tried to show the above by using linear (in-)dependence of $$a_1, \ldots, a_n$$ as an argument. First of all, let's try to linearly combine the above two systems to zero, i.e. check their linear (in-)depence:

\begin{align} & \lambda_1(a_1 - a_n) + \ldots + \lambda_{n-1} (a_{n-1} - a_n) &= 0 & \qquad(1.1)\\ \Leftrightarrow & \lambda_1 a_1 + \ldots + \lambda_{n-1} a_{n-1} - (\lambda_1 + \ldots + \lambda_{n-1}) a_n &= 0 &\qquad(1.2) \end{align}

\begin{align} & \mu_1(a_2 - a_1) + \ldots + \mu_{n-1} (a_{n} - a_1) &= 0 & \qquad(2.1)\\ \Leftrightarrow & -(\mu_1 + \ldots + \mu_{n-1}) a_1 + \mu_1 a_2 + \ldots + \mu_{n-1} a_{n} &= 0 &\qquad(2.2) \end{align}

If $$a_1, \ldots, a_n$$ are linearly independent, we follow from (1.2) and (2.2) that $$\lambda_1 = \ldots = \lambda_{n-1} = 0$$ and $$\mu_1 = \ldots = \mu_{n-1} = 0$$ and thus the above two systems are both linearly independent.

If $$a_1, \ldots, a_n$$ are not linearly independent, then there's a non-trivial solution, i.e. $$\exists i \in \left\{1, \ldots, n\right\}: \kappa_1 a_1 + \ldots \kappa_n a_n = 0,\ \kappa_i \neq 0 \qquad (3)$$

But unfortunately, we can't directly follow that the above two systems must now be linearly dependent. Let's say $$(1, 0, \ldots, 0)$$ is a solution for (3) (and as the solutions of a homogeneous system of equations forms a subvectorspace, I reckon all $$(x, 0 \ldots, 0), x \in \mathbb{F}$$ would be a solution in that case). Then we still wouldn't be able to find a non-trivial solution for (2.2), as $$\kappa_1 = -(\mu_1 + \ldots + \mu_{n-1}) = 0$$.

## The question

So I ask you: is my above proof-idea salvageable? What's a good way to prove the statement to be shown?

Let $$S=\{a_i-a_n\}_{i=1}^{n-1}$$ and $$T=\{a_{j+1}-a_1\}_{j=1}^{n-1}$$. We have to show that S is LI $$\iff$$ T is LI.
Assume that $$S$$ is LI. Let us consider the linear combination of vectors in $$T$$ that adds up to the zero vector. \begin{align*} c_1(a_2-a_1)+c_2(a_3-a_1)+\dotsb+c_{n-1}(a_n-a_1) & =0\\ c_1(a_2-a_{\color{red}{n}})+c_2(a_3-a_{\color{red}{n}})+\dotsb+c_{n-1}(a_n-a_1) & =\left(\sum_{k=1}^{\color{blue}{n-2}}c_i\right)(a_1-a_{\color{red}{n}})\\ c_1(a_2-a_n)+c_2(a_3-a_n)+\dotsb-\left(\sum_{k=1}^{\color{red}{n-1}}c_i\right)(a_1-a_n) & =0 \end{align*} Now we use the fact that the set $$S$$ is LI. This means these coefficients must be zero. Thus we have $$c_1=c_2=\dotsb=c_{n-2}=0 \quad \text{ and } \left(\sum_{k=1}^{n-1}c_i\right)=0.$$ This gives us $$c_i=0$$ for all $$i \in \{1,2,3, \ldots, n-1\}$$. Thus the set $$T$$ is LI.
1. I'm not sure your proof can be easily finished for the case when $$a_1,\dots, a_n$$ are linearly dependent.
2. Let $$b_i:=a_i-a_n$$ (with $$i) and $$c_i:=a_i-a_1$$ (with $$i>1$$).
Then we have $$c_2=b_2-b_1,\ c_3=b_3-b_1,\dots$$ and $$c_n=-b_1$$.
But, this is a linear basis transformation, with inverse $$b_1=-c_n$$ and $$b_i=c_i-c_n$$ for $$i>1$$.