Can someone with access to Shankar's QM book help me (vector spaces)?

AxiomOfChoice
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On pg. 2, Shankar seems to just assume (I'm guessing) that 1|V\rangle = |V\rangle for all vectors |V\rangle when he does the exercise at the bottom of the page. Is this true, or is it possible to prove 1|V\rangle = |V\rangle from the axioms he lists?
 
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I think that this is what the author means by "do what is natural". Usually, however, 1|v> = |v> is just listed as an axiom ("there is a scalar multiplication r|v> with 1|v> = |v> and 0|v> = |0>").
 

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