Proof check that [imath]\text{span}(v_1,\dots,v_i,\dots,v_n) = \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)[/imath]

[Ozma]

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

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

[math]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[/math]
The LHS is a linear combination of the vectors [imath]v_1,\dots, v_i, \dots, v_n[/imath] while the RHS is a linear combination of the vectors [imath]v_1,\dots,\alpha v_i, \dots, v_n[/imath]. So the equality of LHS and RHS implies [imath]\text{span}(v_1,\dots,v_i,\dots,v_n) \subseteq \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)[/imath] and [imath]\text{span}(v_1,\dots,\alpha v_i,\dots,v_n)\subseteq \text{span}(v_1,\dots,v_i,\dots,v_n)[/imath], ending the proof.

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

 

[blamocur]

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

 

[Ozma]

@blamocur: Thanks for the help. The paragraph with [imath]\beta[/imath] was to clarify that the coefficient [imath]\alpha^{-1} c_i[/imath] 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 [imath]\alpha^{-1} c_i[/imath]. 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 [imath]v \in \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)[/imath], hence [imath]v=c_1 v_1 + \dots + c_i (\alpha v_i)+\dots+c_n v_n[/imath] for some [imath]c_1,\dots,c_i,\dots,c_n \in \mathbb{K}[/imath]. Since:
[math]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[/math]with [imath]c_i \alpha \in \mathbb{K}[/imath] because fields are closed with respect to their elements multiplication. This shows that [imath]v[/imath] is a linear combination of [imath]v_1, \dots, v_i, \dots, v_n[/imath] and so [imath]v \in \text{span}(v_1,\dots,v_i,\dots, v_n)[/imath].

Now it should be correct. Do you agree?

 

[blamocur]

@blamocur: Thanks for the help. The paragraph with [imath]\beta[/imath] was to clarify that the coefficient [imath]\alpha^{-1} c_i[/imath] 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 [imath]\alpha^{-1} c_i[/imath]. 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 [imath]v \in \text{span}(v_1,\dots,\alpha v_i,\dots,v_n)[/imath], hence [imath]v=c_1 v_1 + \dots + c_i (\alpha v_i)+\dots+c_n v_n[/imath] for some [imath]c_1,\dots,c_i,\dots,c_n \in \mathbb{K}[/imath]. Since:
[math]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[/math]with [imath]c_i \alpha \in \mathbb{K}[/imath] because fields are closed with respect to their elements multiplication. This shows that [imath]v[/imath] is a linear combination of [imath]v_1, \dots, v_i, \dots, v_n[/imath] and so [imath]v \in \text{span}(v_1,\dots,v_i,\dots, v_n)[/imath].

Now it should be correct. Do you agree?

Agree.