Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that if $r \ne s$, then $u = o$, the zero vector in $V$ and If $u \ne o$, then $r = s$. Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that:

If $r \ne s$, then $u = o$, the zero vector in $V$.
If $u \ne o$, then $r = s$.

EDIT:
Attempt:
I know that this question is a little bit easy because it is a basic concept. But, although so,
I got a little confused about my answer below.
Let $r,s \in \Bbb F$, a field, and $u \in V$. Now,
\begin{align*}
r \cdot u &amp;= s \cdot u \\
r \cdot u - s \cdot u &amp;= o \\
(r-s) \cdot u &amp;= o \\
\end{align*}
What next?
Assume that $V$ is over field, say $F$. Then, if $r \ne s$,
\begin{align*}
(r-s) \cdot (r-s)^{-1} \cdot u &amp;= o \cdot (r-s)^{-1} \\
u &amp;= o,
\end{align*}
and if $u \ne o$, then we must have $r-s = 0$ i.e. $r=s$.
Does it true?
EDIT: You are correct so far, but you must explain why $(r - s) \mathbf u = \mathbf 0$ implies that $r - s = 0$ or $\mathbf u = \mathbf 0.$ Equivalently, we can show that if $r - s \neq 0,$ then $\mathbf u = \mathbf 0.$ Given that $r - s \neq 0$ and $r$ and $s$ are elements of a field, what can you say about $r - s?$ Is there a way to &quot;cancel&quot; $r - s$ from the left-hand side? Once you study bases, you may use the below argument. (It can't hurt to practice.)

Consider a basis $\mathscr B$ of $V.$ Every vector of $V$ can be written uniquely as a finite $\mathbb F$-linear combination of basis vectors of $V.$ Particularly, there exist unique scalars $c_1, \dots, c_n$ and basis vectors $\mathbf v_1, \dots, \mathbf v_n$ such that $\mathbf u = c_1 \mathbf v_1 + \cdots + c_n \mathbf v_n.$ Given that $r \mathbf u = s \mathbf u,$ we have that $$(r c_1) \mathbf v_1 + \cdots + (r c_n) \mathbf v_n = (s c_1) \mathbf v_1 + \cdots + (s c_n) \mathbf v_n.$$ What can you say about the relationship between $r c_i$ and $s c_i$ for each integer $1 \leq i \leq n?$ From there, I believe that you can conclude the desired results.
I don't believe the question as posted is complete - unlike ordinary numbers, vector Dot Products do not have the cancellation property. From Wikipedia article for Dot Product
If $a ⋅ b = a ⋅ c$ and $a  ≠  0 $, then we can write: $ a  ⋅ ( b  −  c ) = 0$ by the Distributive Law; this just means that  $a$  is perpendicular to $( b  −  c )$, which still allows $( b  −  c ) ≠  0 $, and therefore allows $b  ≠  c $.
To put it another way, $r ⋅ u = s ⋅ u$ (with $r ≠ s$) can still be true if $u ⟂ (r - s)$ and $u ≠ 0$.
EDIT: Assuming a downvote means someone doesn't believe me, I'll work through a simple example with three perpendicular, non-zero unit vectors.
Let $r = (1,0,0)$, $s = (0,1,0)$, and $u = (0,0,1)$.
$$r ⋅ u = s ⋅ u$$
$$\left\lVert r \right\rVert \times  \left\lVert u \right\rVert \times \cos \theta_{ru} = \left\lVert s \right\rVert \times \left\lVert u \right\rVert \times \cos \theta_{su}$$
$$1 \times 1 \times \cos \theta_{ru} = 1 \times 1 \times \cos \theta_{su}$$
Obviously, being perpendicular unit vectors, $\theta_{ru} = \theta_{su} = 90^\circ$, so we get
$$1 \times 1 \times \cos 90 = 1 \times 1 \times \cos 90$$
$$1 \times 1 \times 0 = 1 \times 1 \times 0$$
$$0=0$$