I have to decide if the following statement is true or false:
The operation of vector negation is a bijection from the set of free vectors to itself.
Obviously the domain and codomain are correct, but is it a bijection?
I want to say yes, because, although a free vector (and thus its negative) can be represented by an infinite amount of bound vectors, they all have to have the same length and direction, so they represent the same unique free vector. In other words, no matter their positions in 3-space, can all free vectors with the same length and direction be considered one and the same free vector, thus making vector negation a bijection? Thanks in advance!
Let $f:V\rightarrow V$ where $V$ is a vector space such that $f(x)=-x$. We will show that $f$ is a bijection.
Claim: $f$ is one-one.
Proof: $f$ is one-one means that if $f(x)=f(y)$ then $x=y$. If $f(x)=f(y)$ then $-x=-y$ and therefore $x=y$.
Claim: $f$ is onto.
Proof: $f$ is onto means that for all $y\in V$ there exists and $x$ in $V$ such that $f(x)=y$. Consider any $y\in V$, then $f(x)=y$ is equivalent to $-x=y$ and therefore $x=-y$. Does $-y$ exist in $V$? It does because vector spaces are closed under scalar multiplication.
Since $f$ is both one-one and onto, $f$ is a bijection. QED