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In a linear algebra textbook they define a vector space to be a nonempty set $V$ of objects that satisfy certain properties.

One of these properties is that $\forall\vec{u}\in V(1\vec{u}=\vec{u})$

The only way I can think of that property NOT holding would be if scalar multiplication were not defined for some object (not necessarily a vector).

Are there any other cases where multiplication of $n$ by 1 (the scalar multiplicative identity) would not equal $n$?

(not really sure what to tag this with)

scott_fakename
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  • To whomever downvoted: Care to explain? – scott_fakename Nov 02 '13 at 01:00
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    I did not downvote – ILoveMath Nov 02 '13 at 01:01
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    If $1\vec{u}\neq\vec{u}$, then you don't have a vector space. What exactly are you asking? Are you looking for an algebraic structure which does not have $1$ as an identity? – Chuck Franklin Nov 02 '13 at 01:07
  • @austin Exactly so. I was confused, since whenever I have encountered scalar multiplication of anything, multiplication by 1 has always been an identity; Are there cases where this doesn't hold? – scott_fakename Nov 02 '13 at 01:09
  • @anorton Aside from a general philosophy that it would be poor communication to make "1" act in a non-identity way, this example doesn't make scalar multiplication take vectors to vectors (in the same vector space). $r\vec{v}$ should at least be back in the same arena that $\vec{v}$ is in, but here $r\vec{v}\in\mathbb{R}$ while $\vec{v}\in\mathbb{R}^2$. So OP wouldn't be considering maps like this. – 2'5 9'2 Nov 02 '13 at 06:02

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Think of it this way. If you did not explicitly make $1\vec{u}=\vec{u}$ into one of the vector space axioms, you would have no justifiable way to conclude that $1\vec{u}=\vec{u}$; this identity does not follow from the other vector space properties. (You can try to derive it, but you'll never get there using only the other axioms.)

Your intuition is good: you want $1\vec{u}=\vec{u}$ to be true in order to have an abstract vector space be in good correspondence to your perception of physical vector spaces and scaling. So you just need to put this relation $1\vec{u}=\vec{u}$ into the list of axioms.

To answer your question more directly, any mathematical structure that has $1\vec{u}\neq\vec{u}$ is, in my opinion, just being needlessly confusing, using the symbol "$1$" to mean something different from all common meanings of that symbol, or using the multiplication operation to mean something different from all common meanings. (Added later: as MarkS. points out in the comments, concatenation is indeed a common meaning for two items written adjacent to each other that I had overlooked.)

2'5 9'2
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    So you're saying they're defining properties, not necessarily listing restrictions? That makes sense. Thanks! – scott_fakename Nov 02 '13 at 01:10
  • @scott_fakename Yes - you are giving a definition to an abstract thing that tries to be a model for physical counterparts. – 2'5 9'2 Nov 02 '13 at 01:12
  • @alex.jordan What about strings of bits/digits under concatenation? Then $1u\ne u$, and you'd write something like "$eu=u$ for any string $u$", where $e$ stands for the empty string. I'm not saying that's the best way to handle it, but it's tidier than something like $``1"u$. – Mark S. Nov 02 '13 at 01:27
  • @MarkS. Then you would be working in a different field than the real numbers. The use of $1$ in the definition of the vector space is due to fact that $1$ is the multiplicative identity in the field of real numbers; if you are working in a field of strings, then you would indeed have a different definition of "$1$" – apnorton Nov 02 '13 at 02:54
  • @anorton Of course. I suppose I read alex.jordan's message as saying that all cases where $1u\ne u$ involve being needlessly confusing, but it looks like I misinterpreted one of the commas. – Mark S. Nov 02 '13 at 02:58
  • @MarkS. Oh--ok. :) – apnorton Nov 02 '13 at 03:02
  • @MarkS. I think you are right - concatenation is indeed a common meaning for two adjacent items. – 2'5 9'2 Nov 02 '13 at 04:24