$V$ is a vector space over a field $F$ if it satisfies the following conditions.

  1. $(V,+)$ is an abelian group.
  2. $1 \in F $ such that $1.\alpha=\alpha$ for every $\alpha \in V$
  3. $(c_{1}c_{2})\alpha=c_{1}(c_{2}\alpha)$
  4. $(c_{1}+c_{2})\alpha =c_{1}\alpha +c_{2}\alpha$
  5. $c(\alpha+\beta)=c\alpha+c \beta $

I want to find examples of sets which are not vector spaces over some Field $F$ by violating exactly one condition among them.

Examples for violating 1 is easy. But what about second one only ? That means $1 \in F$ does not satisfy $1.\alpha=\alpha$ ?

Here scalars $c,c_{1},c_{2}$ are from $F$ and vectors $\alpha$ and $\beta$ are from $V$

  • 3
    $\begingroup$ For 2. consider what happens if all $cv = 0$ for $c\in F$ and $v\in V$. $\endgroup$ – Tobias Kildetoft Jun 24 '15 at 15:20
  • $\begingroup$ The set of all polynomials with real coefficients of degree $n$ is not a vector space, because is not closed respect sum $(+)$ $\endgroup$ – Luis Felipe Jun 24 '15 at 15:23
  • $\begingroup$ To violate $2$ does not mean that $1 \in F$ does not satisfy $1\cdot a=a$ (this is part of the definition of unity, we use $1$ as a symbol for the element that has that property), but that there is no element of $a_0 \in F$ such that $a_0\cdot a=a$ $\endgroup$ – Lonidard Jun 24 '15 at 15:26
  • $\begingroup$ @bharb It precisely means that $1\cdot a\neq a$ for some $a\in V$ (in $V$, not in $F$). $\endgroup$ – Tobias Kildetoft Jun 24 '15 at 15:30
  • $\begingroup$ Can any one give a concrete example where all other axioms except 2 does not hold. That means for at least one $a \in V$ and $1 \in F ,1.a \neq a$ $\endgroup$ – Madhu Jun 24 '15 at 16:53

Violating only 1:

Be $F = \mathbb F_2$ and $V = \{0,a,b\}$ with the addition table $$\begin{array}{c|ccc} + & 0 & a & b\\ \hline 0 & 0 & a & b\\ a & a & 0 & a\\ b & b & b & 0 \end{array}$$ and the obvious multiplication rule $0\alpha=0$, $1\alpha=\alpha$. Clearly $(V,+)$ is not an abelian group, but one easily verifies that all the other conditions hold.

Violating only 2:

Here I'll take the example proposed by Tobias Kildetoft in the comments:

Be $F$ an arbitrary field, and $(V,+)$ an arbitrary non-trivial abelian group. Decine $c\alpha = 0_V$ for all $c\in F$ and $\alpha \in V$.

Violating only 3:

Take $F=\mathbb C$ and $V=\mathbb R^n$ with the usual per-entry addition (thus fulfilling condition 1). Define $c\alpha = \operatorname{Re}(c)\alpha$ where on the right hand side I've used the usual scalar multiplication with real numbers. Since $\operatorname{Re}(z)$ is linear, conditions 2, 4 and 5 are fulfilled. However if $c_1 = c_2 = \mathrm i$, then the left hand side of condition 3 is $(\mathrm i^2)\alpha = -\alpha$ while the right hand side is $\mathrm i(\mathrm i\alpha) = \mathrm i(\operatorname{Re}(\mathrm i)\alpha) = 0$.

Violating only 4:

Be again $F=\mathbb C$ and $V=\mathbb R^n$ with the usual addition. However now define $c\alpha = \left|c\right|\alpha$.

Violating only 5:

OK, here I'm giving up for now.

  • $\begingroup$ $V={0,a,b} $ with the operation given is not a group at all (associativity $(a+b)+a \neq a+(b+a)$ ! I like this example.Now I can ask the same question with splitting the conditions for abelian group one by one... $\endgroup$ – Madhu Jun 27 '15 at 15:55

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