Vector spaces have been confusing me for a while, I haven't really grasped a clear image of the idea yet.

The question is simple though

Determine whether the set of real numbers (x,y) with operations $$(x_1,y_1)\oplus(x_2,y_2) = (x_1+x_2+1, y_1+y_2+3)$$ $$ k \cdot(x,y) = (kx,y) $$ is a vector space or not. Check all axioms.

The axioms that passed were 1,6,2,3,9 & 10. The axioms that failed were 7,8

The axioms that confused me were 4 & 5.

In axiom 4 the rule is that there should exist $\vec 0$ such that $$ \vec0 + \vec u = \vec u$$

Here's where it confused me, is $\vec 0$ always $= (0,0)$?

Because in this case it wouldn't work, since $$\vec0 + \vec u = (0,0) + (x_1,y_1)$$ $$= (x_1+1, y_1+3)$$ Unless $\vec 0$ = (-1,-3) then this axiom would be true.

But then when I checked axiom 5, where the rule is $$-\vec u + \vec u = \vec 0$$ If we apply it we get $$-\vec u + \vec u = (-x_1,-y_1) + (x_1,y_1)$$ $$= (1,3)$$

So first I don't know if the zero vector could be something different and in this case it doesn't help. Second, in axiom 5, is the negative in $-\vec u$ considered as a scalar and should be treated it as such? So is $-\vec u$ actually $(-x_1, y_1)$ ? Or is it correct as I previously did?

  • 3
    $\begingroup$ Your $\vec 0$ need not be (0,0). $\endgroup$ Apr 15 '18 at 5:52
  • $\begingroup$ @LordSharktheUnknown thank you for clearing that up $\endgroup$
    – Kode Ch
    Apr 15 '18 at 14:29

Axiom 4. For each element $x$ in $V$ there is an element $y$ in $V$ s.t.


In your case $0_V = (-1, -3),$ so if $u=(a,b), v=(-a-2,-b-6)$, clearly


Your error is that you directly said $\large-\vec{u}=-1\color{blue}{\cdot_R}\vec{u},$ then you're using an operation not defined! And even if you correctly use the defined operation,

$$\large-1\cdot \vec{u}=(-x_1,y_1)\color{red}{=_?}-\vec{u},$$

you haven't prove it!

In short, $\Large-\vec{u}$ means nothing more than it's the inverse of $\Large\vec{u}.$


Well first, if you get one axiom to fail then it is not a vector space . Second ,O^--> does not have to be (0,0) ,it can be any vector (pair) with U satisfying U circled addition V=V for every vector (pair) ( .Third given U=(a,b) ,you must find a vector W =(c,d) which ,when added to U (add means using your circled addition). (p,q)+(r,s)=(p+q,r+s) is NOT the addition used for this problem . In the definition and list of axioms for a vector space the addition is always denoted by '+' .but you must change this to circled '+'.


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