Is this $0v=0$ proof correct?

Question

I'm taking my first linear algebra course at university and recently I've been introduced to vector spaces. Now, the teacher has asked us to prove that $0v=0$ for any $v$ using only the definition of a vector space. I've came up with a proof, but I'm unsure whether it's correct or not, as I'm not familiar with proofs in Math.

$$ \begin{align}(1+1)v &= 2v \\ (1+1)v-2v &= 2v-2v\\ [(1+1)-2]v &= 2(v-v) \\ 0v &= 0 \end{align} $$

If this is correct I can go on proving other properties of vector spaces. Any feedback appreciated!

EDIT: Ok, taking into account José's answer, I came up with this one $$ (1+0)v=v \\ 1v+0v=v \\ -v+v+0v=-v+v \\ 0v=0 $$

Answer 1

Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=\bigl((1+1)-2\bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?

Answer 2

You can write a proof that doesn't require $(-1)v=-v$: $$ 0v=(0+0)v\\ 0v=0v+0v\\ 0v-(0v)=0v+0v-(0v)\\ 0=0v $$ Note that $x-y$ actually means $x+(-y)$; with $-(0v)$ I denote the negative of $0v$.

Where does your attempt go wrong? With the same convention as before, you can do $(1+1)v-(2v)=2v-(2v)$. The right hand side is $0$ by definition of $-(2v)$, but you cannot go from $(1+1)v-(2v)$ to $((1+1)-2)v$ unless you prove that $$ -(2v)=(-2)v $$ which is true but unfortunately depends on $0v=0$. Indeed, for any scalar $a$, $$ av+(-a)v=(a+(-a))v=0v=0 $$ so $(-a)v$ summed to $av$ is $0$, which means $(-a)v$ is the negative of $av$: $$ (-a)v=-(av) $$ Note that in all of this the property $1v=v$ has not been used. With it we can also state $$ (-1)v=-(1v)=-v $$

Answer 3

Here another method : where $θ$ is the zero vector and $a$ is a scaler We know that $av$ is a scaled vector and $v + θ = v$ So : $$av + θ = av $$ Hence :$$θ = (a-a)v$$ $$θ = 0v$$