proving identities of a vector space

Question

I am trying to solve the question shown below:

In order to solve it, first I said $u=1u$ (a property of being a vector space). Then, I grouped the sum $u+(-1)u=1u+(-1)u=u(1+(-1))=0u$.

Now, both parts of question require me to solve $0u$=0. I was thinking of maybe using the property: $a(bu)=b(au), u∈ V; a,b∈R$ to prove this identity. Namely, by making $0u=0(1u)=1(0u)$=0.

Is that the right way to do it, or are there better ways to prove it? Thank you very much in advance.

Answer 1

$$0\vec{u}=(0+0)\vec{u}=0\vec{u}+0\vec{u}.$$ Thus, $$-0\vec{u}+0\vec{u}=-0\vec{u}+(0\vec{u}+0\vec{u})$$ or $$\vec{0}=(-0\vec{u}+0\vec{u})+0\vec{u}$$ or $$\vec{0}=\vec{0}+0\vec{u},$$ which gives $$0\vec{u}=\vec{0}.$$

Now, $$\vec{u}+(-1)\vec{u}=1\vec{u}+(-1)\vec{u}=(1+(-1))\vec{u}=0\vec{u}=\vec{0}.$$