In this lecture we define and study some common properties of linear maps, called surjectivity, injectivity and bijectivity.
A map is said to be:
surjective if its range (i.e., the set of values it actually takes) coincides with its codomain (i.e., the set of values it may potentially take);
injective if it maps distinct elements of the domain into distinct elements of the codomain;
bijective if it is both injective and surjective.
Before proceeding, remember that a function between two linear spaces and associates one and only one element of to each element of , and the function is is said to be a linear map (or linear transformation) if and only iffor any two scalars and and any two vectors .
The set is called the domain of , while is the codomain.
Other two important concepts are those of:
null space (or kernel), a subset of the domain defined as:
range (or image), a subset of the codomain defined as:
Both the null space and the range are themselves linear spaces ( subspaces of and respectively).
Let us start with a definition.
Definition Let and be two linear spaces. Let be a linear map. The transformation is said to be surjective if and only if, for every , there exists such that
In other words, every element of can be obtained as a transformation of an element of through the map .
When is surjective, we also often say that is a linear transformation from "onto" .
Since the range of is the set of all the values taken by as varies over the domain, then a linear map is surjective if and only if its range and codomain coincide:
Example Let be the space of all column vectors having real entries. Let be the linear map defined by the matrix product where In order to find the range of , denote by and the two entries of a generic vector , so thatIf you are puzzled by the fact that we have transformed matrix multiplication into a linear combination of columns, you might want to revise the lecture on matrix products and linear combinations. As varies over the space , the scalar can take on any real value. Therefore, the range of is the subspace spanned by the vectorMore formally, we have thatThere are elements of that do not belong to . For example, the vector does not belong to because it is not a multiple of the vector Since the range and the codomain of the map do not coincide, the map is not surjective.
Example If you change the matrix in the previous example tothenwhich is the span of the standard basis of the space of column vectors. In other words, the two vectors span all of . As a consequence,and the map is surjective.
We can now define injectivity.
Definition Let and be two linear spaces. Let be a linear map. The transformation is said to be injective if and only if, for every two vectors such that , we have that
Thus, a map is injective when two distinct vectors in always have two distinct images in .
Injective maps are also often called "one-to-one".
Note that, by an elementary rule of logic, if we take the above implicationand we negate it, we obtain the equivalent implication
Example As in the previous two examples, consider the case of a linear map induced by matrix multiplication. The domain is the space of all column vectors and the codomain is the space of all column vectors. A linear transformation is defined by whereWe can write the matrix product as a linear combination:where and are the two entries of . Thus, the elements of are all the vectors that can be written as linear combinations of the first two vectors of the standard basis of the space . In particular, we have thatAs a consequence, if , the two vectors differ by at least one entry and their transformations through also differ by at least one entry, so that . Thus, the map is injective. Note that is not surjective because, for example, the vectorcannot be obtained as a linear combination of the first two vectors of the standard basis (hence there is at least one element of the codomain that does not belong to the range of ).
Example Modify the function in the previous example by settingso thatSetWe have thatand Therefore, we have found a case in which but . We can conclude that the map is not injective.
We can determine whether a map is injective or not by examining its kernel.
Proposition Let and be two linear spaces. A linear map is injective if and only if its kernel contains only the zero vector, that is,
The kernel of a linear map always includes the zero vector (see the lecture on kernels) becauseSuppose that is injective. Then, there can be no other element such that and Therefore,which proves the "only if" part of the proposition. Now, suppose the kernel contains only the zero vector. Take two vectors such thatThen, by the linearity of we have thatThis implies that the vector belongs to the kernel. But we have assumed that the kernel contains only the zero vector. As a consequence, We have just proved thatAs previously discussed, this implication means that is injective. The latter fact proves the "if" part of the proposition.
We conclude with a definition that needs no further explanations or examples.
Definition Let and be two linear spaces. A linear map is said to be bijective if and only if it is both surjective and injective.
Below you can find some exercises with explained solutions.
As we explained in the lecture on linear maps, a linear function is completely specified by the values taken by on a basis for .
Let be a basis for and be a basis for .
Specify the function as follows:
Is the function surjective?
The vector is a member of the basis . Therefore . Since is a basis for , any element of the domain can be written aswhere and are scalars. Therefore, the elements of the range of take the formIn other words, the elements of the range are those that can be written as linear combinations of and . But cannot be written as a linear combination of and because altogether they form a basis, so that they are linearly independent. Thus, belongs to the codomain of but not to its range. Therefore, codomain and range do not coincide. As a consequence, the function is not surjective.
Determine whether the function defined in the previous exercise is injective.
Suppose are such that . Then, by the uniqueness of the representation in terms of a basis, we have thatwhere are scalars and it cannot be that both and . Therefore,where we assert that the last expression is different from zero because: 1) and because and are members of a basis; 2) it cannot be that both and . Therefore, We have just proved that Therefore is injective.
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