# Prove that if the following two vectors are not parallel that the following holds.

Given $a \vec{x} + b \vec{y}=c \vec{x} + d \vec{y}$

prove that if $\vec{x} ,\vec{y}$ are not parallel then a=c and b=d.

My thought process.

i) First i tried Leaning on the fact that if $\vec{x} ,\vec{y}$ are not parallel must mean that $\vec{x} ,\vec{y}$ are linearly independent for some reason the idea that no linear combination of the two vectors is the zero vector except 0 of both of them is important. i can see this in the case when one is simply a scalar multiple of anther but when they start at different points im a bit lost.

ii) Second idea the concept of vector space's have a list of specific rules that would help a lot but this still leaves the problem of showing that parallel means that all linear combinations of these two vectors does not form a vector space still leaves a loose end.

iii) so i tried going straight to abstract algebra i eventually came to the conclusion that vector spaces were a field with the binary operations of addition and scalar multiplication and simply a commutative ring with unity with respect to the dot product

iv) I have come to the conclusion that what i want to prove to start is that $a_1 \vec{x} + a_2 \vec{y}=0$ iff $a_1=a_2=0$ And that only for some non-zero $a_1 , a_2$ can this equal 0 iff $\vec{x} , \vec{y}$ are parallel.

For example $\vec{x}= \left[ {\begin{array}{cc} 1 \\ 1 \\ \end{array} } \right]$ and $\vec{y}= \left[ {\begin{array}{cc} 2 \\ 2 \\ \end{array} } \right]$ then $a_1=-2 a_2$ will always yield the zero vector even when $a_2 \neq 0$

This seems to be a step in the right direction though it does not prove that for all parallel vectors so i still can't assume the opposite is true for for all non-parallel vectors.

v) My last attempt is to use a logical argument as math seems to be failing me completely here. so lets first assume that $\vec{x} ,\vec{y}$ are perpendicular clearly a scalar multiple of a vector laying entirely on the x axis for example can ever make a vector in the direction of the y axis and any all other perpendicular vectors is simply a rotation of this case. so i we have that $a_1 =a_2 =0$ is the only possible choice that will result in $a_1 \vec{x} = -a_2 \vec{y}$ this logic seems to imply that in the perpendicular case that $a \vec{x} + b \vec{y}=c \vec{x} + d \vec{y}$ or $(a-c) \vec{x} + (b-d) \vec{y}=0$ since (a-c) is simply a scalar and (b-d) is also a scalar clearly a-c= 0 or a=c and (b-d)=0 or b=d showing the desired result.

Well clearly not parallel and perpendicular are not the same thing but if two vectors are not parallel then we should be able to subtract a scalar multiple of the first from the second to remove all of the second vectors x direction leaving it with only a y direction and 0 x axis direction. we can then use this new vector to remove some scalar multiple from the first one to remove all of its y component leaving us with two new vectors that are perpendicular, which i believe implies that the there is no scalar combination of those two vectors that yields zero except both scalars being 0. implying the truth of the last result in this case as well.

Clearly this is not the proper way of doing this and it feels more like a logic argument (perhaps full of many holes) rather than a mathematical proof. Could someone prove this in a more formula capacity?

The first approach is the easiest. The equation you are looking at is equivalent to $$(a-c)\vec x + (b-d) \vec y = 0$$ If $x, y$ are not parallel then they are linearly independent, which, by definition, implies $(a-c)=0=(b-d)$