Hello, I'd like to make a, probably stupid, question regarding the axioms that define a vetor space. Among them, there are the axioms:(adsbygoogle = window.adsbygoogle || []).push({});

λ[itex]\cdot[/itex](μ[itex]\cdot[/itex]X) = (λμ)[itex]\cdot[/itex]Χ (1) and 1[itex]\cdot[/itex]Χ=Χ (2)

for all λ,μ in the field and for all X in the vector space, where 1 is the identity of the multiplication on the field and "[itex]\cdot[/itex]" indicates the multiplication of a scalar with a vector. So, my question is: can axiom (2) derived as a special case of (1) and, therefore, is not really an axiom?

I ask this because, setting λ = μ = 1 in (1), we get:

1[itex]\cdot[/itex](1[itex]\cdot[/itex]X) = 1[itex]\cdot[/itex]Χ

Since X is an abitrary vector, 1[itex]\cdot[/itex]Χ will also be an arbitrary vector, say Y. Then the above equation reads:

1[itex]\cdot[/itex]Y = Y

for all Y in the vector space. But this is "axiom" (2)! I understand that, since both axioms are used in the standard definition of vector space, they should indeed be indepentend, despite the above objection. So what is the flaw in my thought about this?

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# On the axioms of vector space

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