I'm working through a linear algebra text, starting right from the axioms. So far I've understood and proved for myself that, for some vector space $V$ over a field $\mathbb{F}$:

  • the additive identity (zero vector) $\mathbf{0}$ in a vector space is unique
  • the additive inverses in a vector space are unique
  • scalar zero times any vector is the zero vector: $\forall \mathbf{v} \in V: 0 \mathbf{v} = \mathbf{0}$
  • any scalar times the zero vector is the zero vector: $\forall a \in \mathbb{F}: a \mathbf{0} = \mathbf{0}$

However, I'm stuck on the converse of the third statement: suppose $a \mathbf{v} = \mathbf{0}$ and $\mathbf{v} \neq \mathbf{0}$. Show that $a$ must be equal to $0$. In other words, show that the scalar zero is the only element of $\mathbb{F}$ that allows for rule of inference number 3 in the above list.

It seems like such a simple thing but I'm not used to proving super basic statements like this axiomatically. My attempt so far is something like:

$$\begin{align}a\mathbf{v} &= \mathbf{0} \quad &\text{(1)} \\ a \mathbf{v} + \mathbf{0} &= \mathbf{0} \quad &\text{(vector additive identity)} \\ a \mathbf{v} + a \mathbf{v} &= \mathbf{0} \quad &\text{(substitute from 1)} \\ (a + a) \mathbf{v} &= \mathbf{0} \quad &\text{(distributive property)} \\ (2a)\mathbf{v} &= \mathbf{0} \\ (2a)\mathbf{v} &= a \mathbf{v} \quad &\text{(substitute from 1)} \\ \mathrm{``therefore"} \quad 2a &= a \\ a &= 0 \end{align} $$

But I'm not sure I'm "allowed" to do that second-to-last step yet, given the things proved so far. I think it might just be a circular argument. Is it?


I figured it out with some prodding; turned out I had all the pieces in front of me already but didn't realize it. Here it is for completeness:

Suppose $a \in \mathbb{F}$, $\mathbf{v} \in V$, and $a \mathbf{v} = \mathbf{0}$. Either $a = 0$ or $a \neq 0$. In the case where $a \neq 0$:

$$\begin{align} a \mathbf{v} &= \mathbf{0} \\ \frac{1}{a} ( a \mathbf{v} ) &= \frac{1}{a} \mathbf{0} \\ (\frac{1}{a} a) \mathbf{v} & = \mathbf{0} \\ 1\mathbf{v} &= \mathbf{0} \\ \mathbf{v} &= \mathbf{0} \end{align}$$

But then suppose further that $\mathbf{v} \neq \mathbf{0}$. Then $a = 0$ by the contrapositive.

  • $\begingroup$ Your step $\;(2a)v=av\implies 2a=a\;$ would require some serious inspections: you cannot divide by vectors, so how did you "cancel"? $\endgroup$
    – DonAntonio
    Mar 25, 2019 at 22:48
  • 1
    $\begingroup$ Reminds me of this $\endgroup$ Mar 26, 2019 at 0:31

2 Answers 2


An idea for you on the same varation: suppose the scalar $\;a\in\Bbb F\;$ is not zero. It then has an inverse $\;a^{-1}\;$, and:

$$av=0\implies a^{-1}(av)=a^{-1}0\implies 1\cdot v=0\implies v=0$$

The above assume you already know that scalar times the zero vector equals the zero vector...

  • $\begingroup$ Yeah I already proved that part, by that method, but I can't seem to get it to work for nonzero $v$ implying $a=0$. There isn't a corresponding multiplicative inverse for v $\endgroup$
    – dain
    Mar 25, 2019 at 22:54
  • $\begingroup$ If you "already proved that part", then you're done... $\endgroup$
    – amsmath
    Mar 25, 2019 at 23:17
  • $\begingroup$ I don't see how I have? That statement is about the product of a nonzero scalar and the zero vector resulting in the zero vector. But my question is: given that av=0 for nonzero v, show that a is zero. What if there were more than one scalar, distinct from zero, that sent all vectors to the zero vector when multiplied? I mean, I know there isn't, but I don't know how to prove there isn't. $\endgroup$
    – dain
    Mar 25, 2019 at 23:21
  • $\begingroup$ Okay I figured it out, I edited the OP with the answer. I overlooked that it straightforwardly implies a=0 by contrapositive. I went down the wrong track because I thought I had to somehow redo the proof but for scalar zero instead of vector zero. $\endgroup$
    – dain
    Mar 26, 2019 at 0:00

Hint: There was that axiom that said that $1v=v$, was there not?

  • $\begingroup$ Sorry I still don't know where to go with it :s $\endgroup$
    – dain
    Mar 25, 2019 at 23:14
  • $\begingroup$ Okay I figured it out, I edited the OP with the answer. Thanks! $\endgroup$
    – dain
    Mar 26, 2019 at 0:00

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