Vtyjkyuk

Let $x,y$ be any two nonzero vectors in $mathbbR^2$ that are not scalar multiplies of eachother (i.e. are not linearly dependent), and $x-y$ be their difference.

I am wondering why these three vectors, $x,y,x-y$, always form a triangle (subject to repositioning).

I get that it is likely just from the definition of the difference between two vectors, but I don't see how it follows from the definition

I am looking more for a formal proof than intuition.

Edit: By "Form a triangle" I mean that if you draw $x$ and $y$ as starting at the same point, and draw $x-y$ as starting at the tip of $y$, the resulting picture will be a triangle.

For an example see the third picture here

I.e., why is the length of $x-y$ sufficient so that $x-y$, when placed at the tip of $y$, goes from the tip of $x$ to the tip of $y$ (when $x$ and $y$ originate at the same point). How do we know (formally) that there is no "gap"?

In order for three linearly independent vectors $A,B,C $to form a triangle we need to have $A+B+C=0$

Thus $A,B,-A-B$ form a triangle if $A$ and $B$ are independent.

The result could be easily generalized to a polygon.

If you put the vector $x-y$ at the end of $y$ the resulting displacement is $y+(x-y)=x$ This is why they form a triangle.

Sometimes it takes a picture:

By the paralellogram rule for vector addition, $overrightarrowOD$ is $x-y$. It should be clear that $triangleAOB cong triangleOAD$ and that $AB parallel OD$ so $x-y$ indeed “fills the gap.”