# Prove or give a counterexample for linearly independent list of vectors.

Let $$V$$ be a vector space over $$\mathbb{F}$$, and suppose that $$\vec{v}_{1}, \vec{v}_{2}, \dots, \vec{v}_{m} \in V$$ are linearly independent. Prove or disprove: For every choice of non-zero constants $$\alpha_{1}, \alpha_{2}, \dots, \alpha_{m} \in \mathbb{F}$$, the vectors $$\alpha_{1} \vec{v}_{1}, \alpha_{2} \vec{v}_{2}, \dots, \alpha_{m} \vec{v}_{m}$$ are also linearly independent.

Update: the original proof I came up with was too verbose. It's not even good to look at it. Turns out that I over-thought about the problem...

A much simpler :

Suppose there are scalars $$a_1,a_2,\dots,a_m$$ such that the zero vector, $$\textit 0$$, can be written as $$\textit{0}=a_1(\alpha_1v_1)+a_2(\alpha_2v_2)+\cdots+a_m(\alpha_mv_m)$$ Now, this is the same as $$\textit{0}=(a_1\alpha_1)v_1+(a_2\alpha_2)v_2+\cdots+(a_m\alpha_m)v_m$$ and since $$v_1,v_2,\dots,v_m$$ are linearly independent vectors, it follows that $$a_i\alpha_i=0$$ for all $$i=1,2,\dots,m$$. But, $$\alpha_i\neq 0$$, which means $$a_i=0$$. Hence, the vectors $$\alpha_1v_1,\alpha_2v_2,\dots,\alpha_mv_m$$ are also linearly independent.

Consider $$\alpha_1v_1,\alpha_2v_2,\ldots, \alpha_mv_m$$ with $$\alpha_i\neq 0$$ for all $$1\le i\le m$$. Suppose $$\sum_{i=1}^m a_i\alpha_i v_i=0.$$ Since $$v_i$$s are linearly independent, we have $$a_i\alpha_i=0$$. As $$\mathbb F$$ is a field, we can conclude that $$a_i=0$$ for all $$1\le i\le m$$.

Let $$\lambda_1 \alpha_{1} \vec{v}_{1} + \lambda _2 \alpha_{2} \vec{v}_{2}+ \dots \lambda_m \alpha_{m} \vec{v}_{m} =0$$

Since $$\vec{v}_{1}, \vec{v}_{2}, \dots, \vec{v}_{m}$$

are linearly independent, we have

$$\lambda_1 \alpha_{1} = \lambda _2 \alpha_{2}= \dots \lambda_m \alpha_{m} =0$$

Since $$\alpha s$$ are not zero, we have $$\lambda_1 = \lambda _2 = \dots \lambda_m =0$$

That proves the linear independence of $$\alpha_{1} \vec{v}_{1} , \alpha_{2} \vec{v}_{2}, \dots , \alpha_{m} \vec{v}_{m}$$