Let be a linear transformation where
and are vector spaces with scalars coming from the
same field .
is called the domain of and the
codomain. The range of ,
denoted by , is the set
of all possible outputs. Mathematically,
Sometimes, one uses the image of , denoted by
, to refer to the range of .
For example, if is given by for some matrix ,
then the range of is given by the column space of .
Surjection
is said to be surjective (or onto)
if its range equals the codomain. In casual terms, it means that
every vector in can be the output of .
If is surjective, it is called a surjection.
Example
Let be given by
.
Note that every vector in is
of the form since
.
Hence, is not surjective since there are vectors in
(e.g. )
that are not in the range of .
Injection
is said to be injective (or one-to-one)
if for all distinct , .
In casual terms, it means that different inputs lead to different outputs.
If is injective, it is called an injection.
Example
Consider the same in the example above.
It is not injective because for every ,
Bijection
If is both surjective and injective, it is said to be bijective
and we call a bijection.
Testing surjectivity and injectivity
Since is a subspace of ,
one can test surjectivity by testing if the dimension of the range
equals the dimension of provided that is
of finite dimension.
For example, if is given by
for some matrix , is a surjection if and only
if the rank of equals the dimension of the codomain.
(Recall that the rank of gives
the dimension of the column space of .)
To test injectivity, one simply needs to see if the dimension of the
kernel is 0. If it is nonzero, then
the zero vector and at least one nonzero vector
have outputs equal , implying that the linear transformation
is not injective.
Conversely,
assume that has dimension 0 and
take any such that . Then
, implying that .
As has dimension 0 and thus contains only the zero vector,
we must have , implying that
. So no two distinct inputs lead to the same output. Hence,
is injective.
For example, if is given by for some matrix ,
then is an injection if and only if the nullity of is 0.
Example
Let be a linear transformation
given by
Note that
where .
The reduced row-echelon form of is
.
Thus has rank 2 and nullity 1. Since the rank is equal
to the dimension of the codomain , we see from the
above discussion that is surjective. But is not injective
since the nullity of is not zero.
Rank-nullity theorem for linear transformations
The following generalizes the rank-nullity theorem for matrices: