Note: Added my linear algebra book to the Math Books page.
Definition: A function is linear if and for and (or, rather, any field ). Furthermore, it is one-to-one if and only if .
Problem: Let and be vector spaces, and let be linear. Then is one-to-one if and only if carries linearly independent subsets of onto linearly independent subsets of .
Solution: First we will show that if is one-to-one, then it carries linearly independent subsets of onto linearly independent subsets of .
Let be a linearly independent subset of . Suppose . We want to show that is also linearly independent. Suppose that is not. Then there exist not all zero such that
Since is linear, we can rewrite this as
Since and is one-to-one, we must have
But is linearly independent, so this is only possible if , a contradiction. So is linearly independent.
Now, we will prove the converse – if takes linearly independent subsets of to linearly independent subsets of then is one-to-one. Let be nonzero such that . It suffices to show that .
If is linearly dependent, then for some , so
If , we have . In the case that , the set is linearly independent but takes it onto , which is linearly dependent, a contradiction.
If is linearly independent, then is also linearly independent. But since , the set is clearly linearly dependent, a contradiction.
Lastly, we have . So must be a one-to-one function, as desired. QED.
Comment: Basically, we made extensive use of the properties of a linear transformation to prove both directions. A nice corollary to this proof is that if is one-to-one, then is linearly independent if and only if is linearly independent.
Practice Problem: Let and be vector spaces, and let be linear. Then is one-to-one if and only if (here, is the kernal of , defined by ).