So V is not vector over field \Bbb{R}

[Another1]

I don't understand, please ckeck

$$Let$$ $$V=\Bbb{R}^2$$ and $${u=(u_1,u_2), v=(v_1.v_2)}\in\Bbb{R}^2$$ , $${k}\in \Bbb{R}$$ define of operation $$u\oplus v = (u_1+v_1,u_2+v_2)$$ and $$k \odot u =(2ku_1,2ku_2)$$ check V is vector over field $$\Bbb{R}$$ ?
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I think
in a property of the additive inverse $$(-u)+u=0=u+(-u)$$

from define $$k \odot u =(2ku_1,2ku_2)$$
So $$(-1)u = (-1) \odot u =(2(-1)u_1,2(-1)u_2) = (-2u_1,-2u_2) $$
$$(-u)+u = (-2u_1,-2u_2) + (u_1,u_2)\ne 0$$

 

[Olinguito]

Hi Another.

In a vector space $V$ over a field $F$, we need to have for every $v\in V$ that
$$1_F\cdot v\ =\ v$$
where $1_F$ is the multiplicative identity in $F$. Is this condition satisfied for the scalar multiplication $\odot$ in this problem? In other words, is
$$1\odot u\ =\ u$$
true for all $u\in\mathbb R^2$?