Injective and Surjective Linear Transformations
Injective Linear Transformations
Let be a linear transformation. We say that is injective (or one-to-one) if for any vectors in , . The following theorem give us a convenient way to check the injectivity of a linear transformation.
Theorem: is injective if and only if implies .
Proof:
() Suppose is injective. Since we know that , implies because is injective.
() Suppose in such that . So , which implies . In other words, . Hence, by definition, is injective.
The zero vector that satisfies the equation (where is the unknown vector in the equation) is sometimes called the trivial solution. Then the above theorem can be rephrased as follows:
is injective if and only if has only the trivial solution.
This theorem can also be rephrased as follows: Let be the matrix for a linear transformation , where is the column vector. Then is injective if and only if is linearly independent. (Why?)
Surjective Linear Transformation
Let be a linear transformation. We say that is surjective (or onto) if for any vector in , there exists a vector in such that .
As before, let be the matrix for a linear transformation . Then the above definition is equivalent to saying that any vector in is a linear combination of the column vectors of :
In other words, .
For a set of vectors in to span the whole , there must be at least m vectors in the set. Therefore, if a linear transformation is surjective, .
Bijective Linear Transformation
Suppose a linear transformation is both injective and surjective, we say that it is bijective.
Let be the matrix for a linear transformation . Then is bijective if and only if and is linearly independent i.e. is a basis for . Since any basis for must have exactly m vectors, .