Just to be precise here, from my understanding one-to-one is not the same thing as bijective. One-to-one simply means that every vector in the domain of the linear map is mapped to at most one vector in the codomain (if $f: X \to Y$ is a linear map then $f$ is one-to-one if $\forall\ x\ \in X$, there is exactly one $y \in Y$ such that $f(x) = y$).
Bijective implies that a linear map is both one-to-one and "onto" (also called surjective), where "onto" means that the linear map from a domain to a codomain covers every vector in the codomain (for every $y \in Y, \exists x \in X$ such that $f(x) = y$).
Edit: I believe the above actually applies to any function, not just linear functions.
kmcrane
Correct; bijective and one-to-one are not the same thing. One-to-one means "injective," and every bijective map is injective, though not every injective map is bijective (it must also be surjective). We should correct this.
ak-47
So $f(x)=x^2$ is injective, $f(x)=log(x)$ is surjective, and $f(x)=x$ is bijective?
kmcrane
@ak-47: Depends on what your domain is. If $f$ is a map from positive reals to reals, then yes. If it's from reals to reals then no: $x^2$ has two distinct roots for $x \ne 0$, and $\log(x)$ is not defined for $x \leq 0$.
Just to be precise here, from my understanding one-to-one is not the same thing as bijective. One-to-one simply means that every vector in the domain of the linear map is mapped to at most one vector in the codomain (if $f: X \to Y$ is a linear map then $f$ is one-to-one if $\forall\ x\ \in X$, there is exactly one $y \in Y$ such that $f(x) = y$).
Bijective implies that a linear map is both one-to-one and "onto" (also called surjective), where "onto" means that the linear map from a domain to a codomain covers every vector in the codomain (for every $y \in Y, \exists x \in X$ such that $f(x) = y$).
Edit: I believe the above actually applies to any function, not just linear functions.
Correct; bijective and one-to-one are not the same thing. One-to-one means "injective," and every bijective map is injective, though not every injective map is bijective (it must also be surjective). We should correct this.
So $f(x)=x^2$ is injective, $f(x)=log(x)$ is surjective, and $f(x)=x$ is bijective?
@ak-47: Depends on what your domain is. If $f$ is a map from positive reals to reals, then yes. If it's from reals to reals then no: $x^2$ has two distinct roots for $x \ne 0$, and $\log(x)$ is not defined for $x \leq 0$.