If $\lambda a + \mu b = \alpha a + \beta b$ and $a$ and $b$ are not parallel, then $\lambda = \alpha$ and $\mu = \beta$ Going through my textbook, I've come across something I don't understand. it says, if $ \lambda a + \mu b = \alpha a + \beta b $ and the non-zero vectors a and b are not parallel, then $ \lambda = \alpha $ and $\mu = \beta$. Perhaps obvious, but to avoid ambiguity the greek letters are constants. 
So, the reason I don't understand this is because I don't understand how in earth this even works...
 A: We that
$$\lambda a+\mu b=\alpha a+\beta b$$
Thus,
$$(\lambda -\alpha) a+(\mu -\beta)b=0$$
Since $a$ and $b$ are not parallels (linearly independent) , the equality above is true only inte case $\lambda-\alpha=\mu-\beta$. Therefore,
$$\lambda=\alpha\quad\mbox{and}\quad\mu=\beta$$
A: You may know that two vectors are parallel if they are a multiple of one another; that is: (non-zero) vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel if (and only if) there's a number $k \ne 0$ such that $\mathbf{a}=k\mathbf{b}$.
Now, in your case:
$$\lambda \mathbf{a} + \mu \mathbf{b} = \alpha \mathbf{a} + \beta \mathbf{b} \iff \left(\lambda-\alpha\right) \mathbf{a} = \left( \beta - \mu \right)\mathbf{b}$$
Hint: the right-hand side now seems to imply that $\mathbf{a}$ and $\mathbf{b}$ are parallel, unless...
A: Let's rewrite the equality as
$$ \lambda a - \alpha a = \beta b - \mu b$$
or rather as
$$ (\lambda - \alpha)a = (\beta - \mu)b.$$
If $\lambda - \alpha \neq 0$, then this states that
$$ a = \frac{\beta - \mu}{\lambda - \alpha} b,$$
which means that the two vectors are multiples of each other. This is equivalent to $a$ being parallel to $b$. But we know this isn't true!
So $\lambda - \alpha = 0$, i.e. $\lambda = \alpha$. And thus $\beta = \mu$ as well.
