# Prove: If $S$ Linearly Independent So Does $L$

Let $S=\{v_1,v_2,...,v_n\}$ be linearly independent and let $L=\{w_1,w_2,...,w_n\}$ s.t $w_i=v_1+v_i$ where $1\leq i \leq n$

Prove that $L$ is linearly independent

Let assume that $L$ is linearly dependent, WOLG let assume that $w_n$ is linearly dependent

$$w_n=\sum_{k=1}^{n-1}\alpha_kw_k$$

$$v_1+v_n=\sum_{k=1}^{n-1}\alpha_k(v_1+v_k)$$

$$v_1=\sum_{k=1}^{n-1}\alpha_k(v_1+v_k)-v_n=\sum_{k=1}^{n-1}\alpha_kv_1+\sum_{k=1}^{n-1}\alpha_kv_k-v_n$$

In contradiction to the fact that $S$ is linearly independent.

Is this proof valid? is there a way to prove this directly? without contradiction?

## 3 Answers

Suppose $\sum_{i=1}^n a_i w_i =0.$ We need to show that $a_i=0$ for all $i.$

Then $$\left(a_1+\sum_{i=1}^n a_i\right)v_1+\sum_{i=2}^n a_i v_i=0.$$ Since, $S$ is linearly independent, therefore $$a_1+\sum_{i=1}^n a_i=0\\ a_i=0\text{ for } i=2,3,\cdots,n$$ It follows that $a_i=0\text{ for } i=1,2,3,\cdots,n.$ Hence, $L$ is linearly independent.

Your proof is essentially correct, except for some cases such as $\{v_1, v_2\} = \{0, v\}$ with $v \ne 0$. Then $L = \{0, v\}$ is linearly dependent but $v$ cannot be expressed as a linear combination of its predecessors, namely $0$. Also, consider $\{v_1, v_2, v_3\} = \{v, 2v, w\}$ where $v$ and $w$ are linearly independent. Then $L = \{2v, 3v, v + w\}$ which is linearly dependent, but $v + w$ cannot be expressed as a linear combination of $2v$ and $3v$.

You would have to deal with the case $v_1 = 0$ separately (which is easy), and then if $v_1 \ne 0$ you can conclude that there exists $i \in \{2, \ldots, n\}$ such that $$w_i = \sum_{j=1}^{i-1} \alpha_jw_j$$

Alternatively, you can say there exists $i \in \{1, \ldots, n\}$ such that $$w_i = \sum_{j \ne i} \alpha_jw_j$$

It's much cleaner to prove it directly:

Let $\alpha_1, \ldots, \alpha_n$ be scalars such that $\displaystyle\sum_{i=1}^n \alpha_i w_i = 0$. We have:

$$0 = \sum_{i=1}^n \alpha_i w_i = \sum_{i=1}^n \alpha_i (v_1 + v_i) = \left(2\alpha_1 + \sum_{i=2}^n \alpha_i\right) v_1 + \sum_{i=2}^n \alpha_i v_i$$

Since this is a linear combination of the linearly independent set $\{v_1, \ldots, v_n\}$, we conclude that all the scalars are equal to $0$:

$$2\alpha_1 + \sum_{i=2}^n \alpha_i = 0$$ $$\alpha_2 = 0$$ $$\vdots$$ $$\alpha_n = 0$$

So from the first equality we also get $\alpha_1 = 0$. Thus, all scalars $\alpha_1, \ldots, \alpha_n$ are necessarily $0$ so we conclude that $\{w_1, \ldots, w_n\}$ is linearly independent.

The proof you gave is good. You can do this directly. Just use the definition.

Let $a_1, a_2,..,a_n$ be numbers such that $$\sum_{i=1}^n a_i w_i =0$$ That means $$\left(\sum_{i=2}^n a_i +2a_1\right) v_1+ \sum_{i=2}^na_i v_i =0$$ That means $a_2=...=a_n=0$ and $\sum_{i=2}^n a_i +2a_1=0$ so also $a_1=0$. That gives you the required claim.