Suppose $\overrightarrow{X}\in{\bf R}^2$ and $r\in{\bf R}$. Show that if $r\overrightarrow {X}=\overrightarrow{0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$.


Let $\overrightarrow {X}=\left( \begin{matrix} a\\ b\end{matrix} \right)$. Assume $r\neq 0$. Then,

$r\overrightarrow {X}=\overrightarrow {0}$,

$\overrightarrow {X}=\dfrac {1} {r }\overrightarrow {0}$,

$\left( \begin{matrix} a\\ b\end{matrix} \right)=\dfrac {1} {r }\left( \begin{matrix} 0\\ 0\end{matrix} \right)$,

$\left( \begin{matrix} a\\ b\end{matrix} \right)=\left( \begin{matrix} \dfrac {1} {r }0\\ \dfrac {1} {r }0\end{matrix} \right)$,

$\left( \begin{matrix} a\\ b\end{matrix} \right)=\left( \begin{matrix} 0\\ 0\end{matrix} \right)$.

Thus, we have $a=0$ and $b=0$, that is, $\overrightarrow {X}=\overrightarrow {0}$.

Now, assume $\overrightarrow {X}\neq\overrightarrow {0}$. We will show that $r=0$. Then,

$r\overrightarrow {X}=\overrightarrow {0}$,

$\left( \begin{matrix} r a\\ r b\end{matrix} \right)=\left( \begin{matrix} 0\\ 0\end{matrix} \right)$. Since $a\neq 0$ and $b\neq 0$, we obtain $r=0$.

So, we are done.

Can you check my proof?

$$ \newcommand{\Vec}[1]{\overrightarrow{#1}} $$

  • 1
    $\begingroup$ It's true for all vector space. $\endgroup$ – Michael Rozenberg Oct 11 '17 at 14:42
  • 1
    $\begingroup$ Does your problem specify that $X$ has to be a 2 dimensional vector? Or do you have to prove it for any size? $\endgroup$ – Riley Oct 11 '17 at 14:42
  • $\begingroup$ @Riley Yesi right. This is in the plane. $\endgroup$ – pozcukushimatostreet Oct 11 '17 at 14:43
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    $\begingroup$ Your proof looks correct under the assumption that $X\in\mathbb{R}^2$ $\endgroup$ – Riley Oct 11 '17 at 14:49

This is an axiom for vector spaces in general, regardless of vector space dimension.

But if you must prove it, I think there's an easier way, by working in reverse:

Consider both cases.

Let $r=0$

Then $r\vec X=0\vec X=(0+0)\vec X=0\vec X+0\vec X=\color{red}{\vec 0+\vec 0=\vec 0}$

Let $\vec X=\vec 0$

Then $r\vec X=r\vec 0=r(\vec 0+\vec 0)=r\vec 0+r\vec 0=\color{red}{\vec 0+\vec 0=\vec 0}$

The $\color{red}{red}$ is where I used the fact given in the original expression

  • $\begingroup$ This answer is incorrect. You are falsely assuming the conclusion $r=0$ or $\vec{X}=\vec{0}$. $\endgroup$ – Jack Oct 11 '17 at 16:59

Your proof is correct. As long as you have shown that $\vec{X}=\vec{0}$ assuming $r\neq 0$, all the rest of your work are redundant.


This holds in abstracto for any vector space $\mathbf V$ over any field $\Bbb F$, straight from the axioms. For if

$r \vec X = 0 \tag 1$

for $0 \ne r \in \Bbb F$ and $\vec X \in \mathbf V$, then $\exists r^{-1} \in \Bbb F$ and so

$\vec X = 1_{\Bbb F} \vec X = (r^{-1}r)\vec X = r^{-1}(r \vec X) = r^{-1} (0) = 0; \tag 2$

the component representation of $\vec X$ is not needed here.

P S. The OP's proof liiks fine, except one only needs $(a = 0) \vee (b = 0)$ near the end, not $(a = 0) \wedge (b = 0)$.


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