My Linear Algebra textbook offers the following theorem:

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> Let $T: R^n \to R^m$ be a linear transformation. Then $T$ is one-to-one if and only if the equation $T(\vec{x})$ = $\vec{0}$ has only the trivial solution.
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I don't understand why this is true. 

From discrete mathematics, I know that a mapping/function $T: A \to B$ is one-to-one if, for $a \in A$ and $b \in B$, we have that $f(a) = f(b)$ implies that $a = b$. In other words, no two elements in the domain map to the same element in the codomain.

In terms of vectors, that means that no two vectors $\vec{x}$ in the domain $R^n$ map to the same vector $A\vec{x}$ in $R^m$. Or in other words, $A\vec{x} = \vec{b}$ has at most one solution for all $\vec{x} \in R^n$.

In the theorem they give, why are they only talking about $\vec{b} = \vec{0}$, the zero vector? 

What if $T(\vec{x}) = \vec{b}$ has infinitely many solutions for $\vec{b} \ne \vec{0}$, violating the condition for one-to-oneness? 

I don't see why only examining the special case of $T(\vec{x}) = \vec{0}$ can allow us to conclude that the entire mapping is one to one.

**Edit:** or is that not what they're saying? Is it because of the wording "if and only if"? Does the theorem actually translate to:

* if $T$ is one to one, then $T(\vec{x}) = \vec{0}$ has only the trivial solution
* if $T(\vec{x}) = \vec{0}$ only has the trivial solution, then $T(\vec{x})$ is one to one

I can see both of these being true.