Non-zero vector equation that requires you to find a scalar I have stumbled across an exercise that goes like this:

Let vector a be a non-zero vector. Find the value of scalar $k$, given the equation: $(k+1)(k-2)a+ka=2a$

How is it done?
 A: All the terms are some scalar multiple of $a$, so it is possible to divide $a$ out (since it is non-zero):
$$(k+1)(k-2)+k=2$$
Now you have just a simple quadratic equation to solve.
A: Let $V$ be a vector space over a field $F$. 
 Then, for any $\alpha \in F$ and for any $v \in V$, the following holds: 
$\alpha v = \mathbf{0}$ if and only if either $\alpha = 0$ or $v = \mathbf{0}$. 
Proof: 

If $\alpha = 0$ or $v = \mathbf{0}$, then we have $\alpha v = \mathbf{0}$, by the axioms of a vector space. So let's conversely assume that $\alpha v = \mathbf{0}$. If $\alpha = 0$, then we are done. So let's assume that $\alpha \neq 0$. Then there exists an element $\alpha^{-1} \in F$ such that $$ \alpha \alpha^{-1} = 1 =  \alpha^{-1} \alpha. $$
  Now as we have assumed that  $\alpha v = \mathbf{0}$, so we also have 
  $$\mathbf{0} = \alpha^{-1} \mathbf{0} =  \alpha^{-1} ( \alpha v) = \left( \alpha^{-1} \alpha \right) v = 1v = v. $$
  That is, $v = \mathbf{0}$. 

Using the above we can also conclude that, for any  $\lambda, \mu \in F$ and for any non-zero $v \in V$, the following holds: $\lambda v = \mu v$ if and only if $\lambda = \mu$. 
Proof: 

We see that $\lambda v = \mu v$ if and only if $\lambda v - \mu v = \mathbf{0}$, which is the same as $(\lambda - \mu) v = \mathbf{0}$, and the last equation holds if and only if $\lambda - \mu = 0$ (because $v$ is non-zero), and this holds if and only if $\lambda = \mu$. 

In what has gone so far, I've assumed that you are familiar with the definiion of a field and that of a vector space over a field. 
Hope this helps. 
