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For a vector space $V$ over a field $F$ one of the axioms is that for any vector $v\in V$, we have $1v=v$. My intuition told me that this should follow from the other axioms, or more specifically the axiom $a(bv)=(ab)v$. The way I tried to show it was that $1(av)=(1a)v=av$ so that $1u=u$ for any vector $u$ that can be written as $av$ for some other vector $v$ and a scalar $a$. Of course $u$ can be written as $1u$ if we assume the axiom, but not necessarily otherwise.

I suppose this axiom for a vector space does not follow from the other axioms, because then it would be redundant. My question however is then, if there is an example of a "pseudo" vector space, where all the usual axioms hold, but not $1v=v$.

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    $\begingroup$ Suppose $a\vec v=\vec 0$ for all scalars $a$. $\endgroup$
    – lulu
    Sep 3, 2021 at 15:52

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Simply define $av=0$ for all $a\in F$, $v\in V$.

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  • $\begingroup$ Now that I think about it, I should have at least considered the trivial example. Thank you! $\endgroup$
    – spinosarus123
    Sep 3, 2021 at 15:53

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