Proof check that span(v1,…,vi,…,vn)=span(v1,…,αvi,…,vn)\text{span}(v_1,\dots,v_i,\dots,v_n) = \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)

[Ozma]

Let $\mathbb{K}$ be a field, let $V$ be a vector space over $\mathbb{K}$, let $n \in \mathbb{N}\setminus\{0\}$ and let $\alpha \in \mathbb{K}\setminus\{0\}$. Prove that $\text{span}(v_1,\dots,v_i,\dots,v_n) = \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)$ for each $i \in \{1,\dots,n\}$.

My attempt: let $i \in \{1,\dots,n\}$ be arbitrary. Let By hypothesis, $\alpha \ne 0$ and so there exists $\alpha^{-1} \in \mathbb{K}$. Since $\mathbb{K}$ is a field, for each $\beta \in \mathbb{K}$ we have $\beta=1_\mathbb{K} \beta = (\alpha^{-1} \alpha) \beta = (\alpha^{-1}\beta)\alpha$ with $\alpha^{-1} \beta \in \mathbb{K}$ due to the closure properties of fields. So, for elements $v_1,\dots,v_n$ of $V$ and coefficients $c_1,\dots,c_n$ of $\mathbb{K}$, we have:

$$c_1 v_1+\dots+c_i v_i+\dots + c_n v_n = c_1 v_1 + \dots +(\alpha^{-1}c_i)(\alpha v_i) + \dots + c_n v_n$$
The LHS is a linear combination of the vectors $v_1,\dots, v_i, \dots, v_n$ while the RHS is a linear combination of the vectors $v_1,\dots,\alpha v_i, \dots, v_n$. So the equality of LHS and RHS implies $\text{span}(v_1,\dots,v_i,\dots,v_n) \subseteq \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)$ and $\text{span}(v_1,\dots,\alpha v_i,\dots,v_n)\subseteq \text{span}(v_1,\dots,v_i,\dots,v_n)$, ending the proof.

Question: is my proof correct? If not, can someone point out possible mistakes?

 

[blamocur]

Why do you need that paragraph about $\beta$?
Also, I can see how the equality implies the first inclusion but not the second.

 

[Ozma]

@blamocur: Thanks for the help. The paragraph with $\beta$ was to clarify that the coefficient $\alpha^{-1} c_i$ obtained later in the proof was indeed an element of the field, but you are right: I can avoid that by just pointing that out when I introduce $\alpha^{-1} c_i$. You are also right that the second inclusion does not follow from the equality I wrote, I will retry to prove the second inclusion as follow.

Let $v \in \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)$, hence $v=c_1 v_1 + \dots + c_i (\alpha v_i)+\dots+c_n v_n$ for some $c_1,\dots,c_i,\dots,c_n \in \mathbb{K}$. Since:
$$c_1 v_1 + \dots + c_i (\alpha v_i)+\dots+c_n v_n = c_1 v_1 + \dots + (c_i \alpha) v_i+\dots+c_n v_n$$with $c_i \alpha \in \mathbb{K}$ because fields are closed with respect to their elements multiplication. This shows that $v$ is a linear combination of $v_1, \dots, v_i, \dots, v_n$ and so $v \in \text{span}(v_1,\dots,v_i,\dots, v_n)$.

Now it should be correct. Do you agree?

 

[blamocur]

@blamocur: Thanks for the help. The paragraph with $\beta$ was to clarify that the coefficient $\alpha^{-1} c_i$ obtained later in the proof was indeed an element of the field, but you are right: I can avoid that by just pointing that out when I introduce $\alpha^{-1} c_i$. You are also right that the second inclusion does not follow from the equality I wrote, I will retry to prove the second inclusion as follow.

Let $v \in \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)$, hence $v=c_1 v_1 + \dots + c_i (\alpha v_i)+\dots+c_n v_n$ for some $c_1,\dots,c_i,\dots,c_n \in \mathbb{K}$. Since:
$$c_1 v_1 + \dots + c_i (\alpha v_i)+\dots+c_n v_n = c_1 v_1 + \dots + (c_i \alpha) v_i+\dots+c_n v_n$$with $c_i \alpha \in \mathbb{K}$ because fields are closed with respect to their elements multiplication. This shows that $v$ is a linear combination of $v_1, \dots, v_i, \dots, v_n$ and so $v \in \text{span}(v_1,\dots,v_i,\dots, v_n)$.

Now it should be correct. Do you agree?

Agree.