1
$\begingroup$

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...

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.