If $\lambda \not= 0$ and $\lambda x = 0$, then $x = 0$

Question

I am trying to use the definition of vector spaces to prove that, if $\lambda \not= 0$ and $\lambda x = 0$, then $x = 0$.

One proof I have seen begins as follows:

$\lambda 0 = 0$ for each $\lambda$ since $\lambda 0 = \lambda(0, + 0) = \lambda 0 + \lambda 0$.

It then proceeds as follows:

Therefore, since $\lambda \not= 0$,

$$\begin{align} \lambda x = 0 &\Rightarrow \lambda^{-1} (\lambda x) = \lambda^{-1} 0 \\ &\Rightarrow (\lambda^{-1} \lambda) x = 0 \\ &\Rightarrow 1x = 0 \\ &\Rightarrow x = 0 \end{align}$$

It is this part that I am confused about:

$\lambda 0 = 0$ for each $\lambda$ since $\lambda 0 = \lambda(0, + 0) = \lambda 0 + \lambda 0$.

It is not clear to me what purpose this has in the proof, nor is it clear to me that it is even valid (that is, it is not clear to me that it is a valid claim, given the definition of a vector space). This part seems out-of-place to me. I would greatly appreciate it if people would please take the time to explain this.

Answer 1

Read the main proof again:

$$ \begin{align} \lambda x=0 &\Rightarrow\lambda^{-1}(\lambda x)=\color{red}{\lambda^{-1}0}\\ &\Rightarrow(\lambda^{-1}\lambda)x=\color{red}{0}\\ &\Rightarrow\cdots\end{align} $$

One needs to justify that $\lambda^{-1}0=0$. So, it suffices to prove that $\lambda0=0$ for every scalar $\lambda$. I think this is the purpose of the leading line of the proof.

Answer 2

To be completely rigorous: \begin{align*} \lambda\vec{0} = \lambda(\vec{0} + \vec{0}) &\implies \lambda\vec{0} = \lambda\vec{0} + \lambda\vec{0} \\ &\implies (-\lambda)\vec{0} + \lambda\vec{0} = (-\lambda)\vec{0} + (\lambda\vec{0} + \lambda\vec{0}) \\ &\implies ((-\lambda) + \lambda )\vec{0} = ((-\lambda )\vec{0} + \lambda \vec{0}) + \lambda\vec{0} \\ &\implies ((-\lambda) + \lambda )\vec{0} = (((-\lambda) + \lambda )\vec{0}) + \lambda\vec{0} \\ &\implies (0)\vec{0} = ((0)\vec{0}) + \lambda\vec{0} \\ &\implies \vec{0} = \vec{0} + \lambda\vec{0} \\ &\implies \vec{0} = \lambda\vec{0} \end{align*}

Answer 3

It is not a part of the proof, it is the opposite implication, that is, if $x$ is the zero vector, then $\lambda x=0$ for every $\lambda$. The proof of this implication relies on the additive group structure of a vector space. Since the zero vector is neutral with respect to addition, you have that $x=x+0$ for every vector $x$, in particular, for $x=0$, you have that $0=0+0$. Then $\lambda(0+0)=\lambda 0+\lambda 0$ comes from the axioms of a vector space. Once you have the equality $\lambda 0=\lambda 0+\lambda 0$ just use cancellation law, which holds in a group, that is add to both sides of the equality the vector $-\lambda 0$. Or, if you want to go directly, you can write

\begin{align*} 0&=-\lambda 0+\lambda 0\nonumber\\ &=-\lambda 0+\lambda (0+0)\nonumber\\ &=-\lambda 0+(\lambda 0+\lambda 0)\nonumber\\ &=(-\lambda 0+\lambda 0)+\lambda 0\nonumber\\ &=0+\lambda 0\nonumber\\ &=\lambda 0\nonumber \end{align*}