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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?

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Note: You tried to argue by applying the definition of LI to both the set of vectors together and that is why you are unable to complete the proof.

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.

We can reverse this argument ot get the other implication as well.

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  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 $i<n$) 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 $i>1$.
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