Proving something about vector spaces that seems very trivial From Linear Algebra Done Right, Sheldon Axler; p. 19:

Prove that if $a\in \mathbb F,v\in V$ and $av=0$, then $a=0$ or $v=0$.

Here F denotes the vector space of real and complex numbers and V is any vector space over $\mathbb F$. This means that $V$ can be denoted by $\{x_1,x_2,...x_n\}, x_j\in \mathbb F$ for $j=1,2,3,...n$. I have proved this for the case when $a=0$. However the proof when $v=0$
seems incorrect to me.
Pg 8 of the same book says:

Another type of multiplication, called scalar
multiplication, will be central to our subject. Specifically, we need to
define what it means to multiply an element of $F^n$ by an element of F.
We make the obvious definition, performing the multiplication in each
coordinate:
$a(x_1, . . . , x_n) = (ax_1, . . . , ax_n);$. Here $a\in F$ and $(x_1,...x_n)\in F^n$

Coming back to the question, $av=(ax_1,ax_2 . . . , ax_n)$. If $a \neq 0$, it seems obvious to me that $v$ has to be zero (i.e all $x$ have to be zero), because that's the only way the product can be zero. However all the solutions given online are much longer than mine. What am I missing?
P.S: I have just graduated from high school, and this is my first attempt at understanding undergraduate level maths.
 A: Sure, it's obvious that if $a\ne0$, then $(ax_1,ax_2,\ldots,ax_n)=(0,0,\ldots,0)$ if and only if each $x_n$ is $0$. But the question is not about $F^n$. It's about an arbitrary vector space, and then you cannot just say that it's obvious. But you can say that, if $a\ne0$ and $av=0$, then$$0=a^{-1}(av)=(a^{-1}a)v=v.$$
A: If $V=F^n$, then indeed, the proof is very short. In this case, you know that
$$av=(av_1,av_2,\dots, av_n)=(0,0,\dots)$$ and so, because $a\neq 0$, you know that $v_i=0$ for all values of $0$ and therefore, $v=0$.

However, the statement does not allow you to assume that $V=F^n$. The statement only requires that you assume that $V$ is a vector space over $F$. Therefore, because there exist vector spaces ovef $F$ that are not $F^n$, the statement is a stronger statement, and it makes sense that it will be harder to prove.

This is something you should internalize as soon as possible if you want to do well in mathematics:

All you are allowed to assume when proving a statement are the things given in the statement .

So, in your example, you can assume the following:

*

*$F$ is a field, and $a$ is an element of the field

*$V$ is a vector space over $F$, and $v$ is a vector in $V$.

*$a\cdot v =0$
and your proof must follow from these two assumptions (and anything that follows from them, of course).
Now, once you split the case and are looking at the case $a\neq 0$, you can do some other things. You now have (while looking at this case) the additional assumption (4: $a\neq 0$). You can now,for example, start with knowing that $$av=0$$
which you know from (3).
Well, you also know that $a$ is a nonzero element of $F$ (4), so you know it has an inverse (because (1) $F$ is a field and in a field, all nonzero elements are invertible). So, you can multiply the equation above with $a^{-1}$. And you get
$$a^{-1}(av)=a^{-1}0$$
Now, you can use some other things you know. Remember all of them must stem from the three points written above! For example, you could use the fact that in a vector space, you always have
$$\alpha(\beta u) = (\alpha\beta)(u)$$
for any two scalars $\alpha,\beta\in F$ and any vector $u\in V$. And you could also use the fact that in a vector space, for any vector $u$, you know that $$0\cdot u = 0$$
and $$1\cdot u = u.$$
This should already be enough to get you to where you need to be.
