# Vector space basics: scalar times a nonzero vector = zero implies scalar = zero?

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?

EDIT:

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.

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

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...

• 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
– dain
Mar 25, 2019 at 22:54
• If you "already proved that part", then you're done... Mar 25, 2019 at 23:17
• 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.
– dain
Mar 25, 2019 at 23:21
• 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.
– dain
Mar 26, 2019 at 0:00

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

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