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

Question

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...

Answer 1

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.

Answer 2

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 $.

Answer 3

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