Does $\lambda$ $\mathbf u$ $= 0$ imply $\lambda = 0$ or $\mathbf u$ $= 0$? Does $\lambda \mathbf u = 0$  imply $\lambda = 0 $ or $\mathbf u= 0$
where $\mathbf u$ is in a vector space $V$ over a field $F$. 
The case for the vector equalling zero is trivial due to the fact field axioms include a multiplicative inverse, so how could I show lambda is 0?
 A: Yes, for suppose $\lambda \neq 0$. Then
$$\lambda u = 0$$ 
$$\implies u=\frac{1}{\lambda}\lambda u = \frac{1}{\lambda }0 = 0$$
A: This is true as it is actually a theorem about general vector spaces. The proof for this is pretty straightforward as well, by showing that if $\lambda u = \vec{0} $ and $\lambda \ne 0$ then a logical consequence is that $u$ must be the zero vector. And of course, if $\lambda = 0$, it is another known theorem that for all $u$ in a given vector space that $0u = \vec{0}$ ($\vec{0}$ being the zero vector).
A: From the other answers is clear that if $\mathbf v=\mathbf 0_V$ then $\lambda\mathbf v=\mathbf 0_V$ for any $\lambda\in\Bbb F$. Then I will show the other cases.

First I will show that $0_{\Bbb F}\mathbf v=\mathbf 0_V$ for any $\mathbf v\in V$. Suppose that $\mathbf v\neq \mathbf 0_V$, then by the properties of the field $\Bbb F$ and some axioms of a vector space we have that
$$0_{\Bbb F}\mathbf v=(0_{\Bbb F}+0_{\Bbb F})\mathbf v=0_{\Bbb F}\mathbf v+0_{\Bbb F}\mathbf v\tag{1}$$ 
And 
$$0_{\Bbb F}\mathbf v+0_{\Bbb F}(-\mathbf v)=\mathbf 0_V\tag{2}$$
Then adding in both sides of $(1)$ the vector $0_{\Bbb F}(-\mathbf v)$ we can conclude that $0_{\Bbb F}\mathbf v=\mathbf 0_V$.

It remains to show that if $\lambda\neq 0_{\Bbb F}$ and $\mathbf v\neq \mathbf 0_V$ then $\lambda\mathbf v\neq\mathbf 0_V$. 
Suppose to the contrary that $\lambda\neq0_{\Bbb F}$, $\mathbf v\neq \mathbf 0_V$ but $\lambda\mathbf v=\mathbf 0_V$. Then from the axioms of a vector space we knows that $\mathbf v=1_{\Bbb F}\cdot\mathbf v$, hence
$$(\lambda+1_{\Bbb F})\mathbf v=\lambda\mathbf v+\mathbf v=\mathbf v\tag{3}$$ 
Now adding in both sides of $(3)$ the vector $(\lambda+1_{\Bbb F})(-\mathbf v)$ we get
$$(\lambda+1_{\Bbb F})\mathbf 0_V=\lambda(-\mathbf v)\tag{4}$$
and because $\lambda\neq 0_{\Bbb F}$ (and $\lambda\mathbf 0_V=\mathbf 0_V$ as other answers shown) then multiplying both sides of $(4)$ by $1/\lambda$ we get
$$\frac{\lambda+1_{\Bbb F}}{\lambda}\mathbf 0_V=\mathbf 0_V=(-\mathbf v)\implies\mathbf v=\mathbf 0_V\tag{5}$$
but we says that $\mathbf v\neq \mathbf 0_V$, so we get a contradiction. Hence if $\lambda\neq 0_{\Bbb F}$ and $\mathbf v\neq \mathbf 0_V$ then it cannot be possible that $\lambda\mathbf v= \mathbf 0_V$.$\square$
A: I know I already answered but to help you with your confusion I am going to throw in another one.
Let's assume we have a vector space $\mathbf{V}$ and for $\mathbf{u} \in \mathbf{V}$ and $\lambda \in \mathbb{F}$ we know that  $\lambda \mathbf u = \mathbf{0}$ and $\mathbf u \ne \mathbf{0}.$ The next part of the proof could vary as it depends on how the scalar multiplication is defined for $\mathbf{V}$ but for simplicity reasons I am going to let $\mathbf{V} = \mathbb{R}^n$ and show that it works in this case. You can feel free to check for other vector spaces.
Anyway, now we have that $\lambda \mathbf{u} =\lambda \left(u_1, u_2, \dots u_n \right) =\left(\lambda u_1,\lambda u_2, \dots \lambda u_n \right) $ where $u_i \in \mathbb{R}$ for all integers $1 \le i \le n$ and $\exists{u_i}$ such that $u_i \ne 0$. Substituting back in we get $\left(\lambda u_1,\lambda u_2, \dots \lambda u_n \right) = \mathbf{0}$.
Now we have a system of linear equations to solve for. $$\left\{ 
\begin{array}{c}
u_1 \lambda = 0 \\ 
u_2 \lambda = 0 \\ 
\vdots \\
u_n \lambda = 0
\end{array}
\right.$$
We know that $\lambda = 0$ is a solution just from observation. And because of the fact that there is at least one $u_i$ that is not $0$ then it turns out that $\lambda = 0$ is the only solution.
I hope this helps on your confusion.
