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My book lists ten axioms that must hold for a set of objects (vectors) $V$ to be called a vector space. One of those axioms is:

$$1\vec{u} = \vec{u}$$

Is there a reason why this axiom must be on the list? What is its purpose, because it seems kind of obvious? Is there a case when a real scalar, 1, multiplied by a vector doesn't return that same vector?

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    $\begingroup$ It is not that obvious as you think. Try to prove it using the other axioms. $\endgroup$ – NECing Oct 25 '14 at 2:52
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Yes, this axiom is necessary.

Given any vector space $(\mathbb{V}, +, \,\cdot\,)$ over a field $\mathbb{F}$, we can produce another algebraic structure $(\mathbb{V}, +, \odot)$ with "multiplication" map $$\odot: \mathbb{F} \times \mathbb{V} \to \mathbb{V}$$ the zero map $$f \odot \mathbf{u} := \mathbf{0}.$$ Then, $(\mathbb{V}, +, \odot)$ satisfies all of the vector space axioms except $$1 \odot \mathbf{u} = \mathbf{u}.$$

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  • $\begingroup$ This thought popped right into mind. $\endgroup$ – ncmathsadist Oct 25 '14 at 17:06
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Let $V$ be the abelian group $F^n$ where $F$ Is a field. Define $av:=0$ for every $a\in F, v\in V$.

You can show that all the vector space axioms hold except the one you speak of.

The main reason you want 1 to act as the identity is so that you can identify $F$ as a subring of the ring of linear transformations of $V$, sharing the same identity.

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In general, vector space $V$ can be over an arbitrary field $F$. And we can define a scalar multiplication between the element of $V$ and element of $F$.

$1\vec{u} = \vec{u}$ means the multiplicative identity in the field is also the identity of this scalar multiplication.

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