Isomorphisms and homomorphisms both preserve structure. The difference is that homomorphisms needn't be onto and needn't be one-to-one. This means that homomorphisms are a more general kind of map, subject to fewer restrictions than isomorphisms. We will examine what can happen with homomorphisms that is prevented by the extra restrictions satisfied by isomorphisms.
We first consider the effect of dropping the onto requirement, of not requiring as part of the definition that a homomorphism be onto its codomain. For instance, the injection map
is not an isomorphism because it is not onto. Of course, being a function, a homomorphism is onto some set, namely its range; the map is onto the -plane subset of .
- Lemma 2.1
Under a homomorphism, the image of any subspace of the domain is a subspace of the codomain. In particular, the image of the entire space, the range of the homomorphism, is a subspace of the codomain.
Let be linear and let be a subspace of the domain . The image is a subset of the codomain . It is nonempty because is nonempty and thus to show that is a subspace of we need only show that it is closed under linear combinations of two vectors. If and are members of then is also a member of because it is the image of from .
- Definition 2.2
The rangespace of a homomorphism is
sometimes denoted . The dimension of the rangespace is the map's rank.
(We shall soon see the connection between the rank of a map and the rank of a matrix.)
- Example 2.3
Recall that the derivative map given by is linear. The rangespace is the set of quadratic polynomials . Thus, the rank of this map is three.
- Example 2.4
With this homomorphism
an image vector in the range can have any constant term, must have an coefficient of zero, and must have the same coefficient of as of . That is, the rangespace is and so the rank is two.
The prior result shows that, in passing from the definition of isomorphism to the more general definition of homomorphism, omitting the "onto" requirement doesn't make an essential difference. Any homomorphism is onto its rangespace.
However, omitting the "one-to-one" condition does make a difference. A homomorphism may have many elements of the domain that map to one element of the codomain. Below is a "bean " sketch of a many-to-one map between sets. It shows three elements of the codomain that are each the image of many members of the domain.
Recall that for any function , the set of elements of that are mapped to is the inverse image . Above, the three sets of many elements on the left are inverse images.
- Example 2.5
Consider the projection
which is a homomorphism that is many-to-one. In this instance, an inverse image set is a vertical line of vectors in the domain.
- Example 2.6
is also many-to-one; for a fixed , the inverse image
is the set of plane vectors whose components add to .
The above examples have only to do with the fact that we are considering functions, specifically, many-to-one functions. They show the inverse images as sets of vectors that are related to the image vector . But these are more than just arbitrary functions, they are homomorphisms; what do the two preservation conditions say about the relationships?
In generalizing from isomorphisms to homomorphisms by dropping the one-to-one condition, we lose the property that we've stated intuitively as: the domain is "the same as" the range. That is, we lose that the domain corresponds perfectly to the range in a one-vector-by-one-vector way.
What we shall keep, as the examples below illustrate, is that a homomorphism describes a way in which the domain is "like", or "analogous to", the range.
- Example 2.7
We think of as being like , except that vectors have an extra component. That is, we think of the vector with components , , and as like the vector with components and . In defining the projection map , we make precise which members of the domain we are thinking of as related to which members of the codomain.
Understanding in what way the preservation conditions in the definition of homomorphism show that the domain elements are like the codomain elements is easiest if we draw as the -plane inside of . (Of course, is a set of two-tall vectors while the -plane is a set of three-tall vectors with a third component of zero, but there is an obvious correspondence.) Then, is the "shadow" of in the plane and the preservation of addition property says that
Briefly, the shadow of a sum equals the sum of the shadows . (Preservation of scalar multiplication has a similar interpretation.)
Redrawing by separating the two spaces, moving the codomain to the right, gives an uglier picture but one that is more faithful to the "bean" sketch.
Again in this drawing, the vectors that map to lie in the domain in a vertical line (only one such vector is shown, in gray). Call any such member of this inverse image a " vector". Similarly, there is a vertical line of " vectors" and a vertical line of " vectors". Now, has the property that if and then . This says that the vector classes add, in the sense that any vector plus any vector equals a vector, (A similar statement holds about the classes under scalar multiplication.)
Thus, although the two spaces and are not isomorphic, describes a way in which they are alike: vectors in add as do the associated vectors in — vectors add as their shadows add.
- Example 2.8
A homomorphism can be used to express an analogy between spaces that is more subtle than the prior one. For the map
from Example 2.6 fix two numbers in the range . A that maps to has components that add to , that is, the inverse image is the set of vectors with endpoint on the diagonal line . Call these the " vectors". Similarly, we have the " vectors" and the " vectors". Then the addition preservation property says that
|a " vector"
||a " vector"
||a " vector".
Restated, if a vector is added to a vector then the result is mapped by to a vector. Briefly, the image of a sum is the sum of the images. Even more briefly, . (The preservation of scalar multiplication condition has a similar restatement.)
- Example 2.9
The inverse images can be structures other than lines. For the linear map
the inverse image sets are planes , , etc., perpendicular to the -axis.
We won't describe how every homomorphism that we will use
is an analogy because the formal
sense that we make of "alike in that ..." is
"a homomorphism exists such that ...".
Nonetheless, the idea that a homomorphism between two spaces expresses how
the domain's vectors fall into classes that act like the
range's vectors is a good way to view homomorphisms.
Another reason that we won't treat all of the homomorphisms that
we see as above is that many vector spaces are hard to draw
(e.g., a space of polynomials).
However, there is nothing bad about gaining insights from those spaces
that we are able to draw,
especially when those insights extend to all vector spaces.
We derive two such insights from the three examples
2.7 , 2.8,
First, in all three examples, the inverse images are lines or planes, that is,
In particular, the inverse image of the range's zero vector is a line or plane
through the origin— a subspace of the domain.
- Lemma 2.10
For any homomorphism, the inverse image of a subspace of the range is a subspace of the domain. In particular, the inverse image of the trivial subspace of the range is a subspace of the domain.
Let be a homomorphism and let be a subspace of the rangespace . Consider , the inverse image of the set . It is nonempty because it contains , since , which is an element , as is a subspace. To show that is closed under linear combinations, let and be elements, so that and are elements of , and then is also in the inverse image because is a member of the subspace .
- Definition 2.11
The nullspace or kernel of a linear map is
the inverse image of
The dimension of the nullspace is the map's nullity.
- Example 2.12
The map from Example 2.3 has this nullspace
- Example 2.13
The map from Example 2.4 has this nullspace.
Now for the second insight from the above pictures. In
Example 2.7, each of the vertical lines is
squashed down to a single point— , in passing
from the domain to the range, takes all of these one-dimensional
vertical lines and "zeroes them out", leaving the range one dimension
smaller than the domain. Similarly, in
Example 2.8, the two-dimensional domain
is mapped to a one-dimensional range by breaking the domain into lines
(here, they are diagonal lines), and compressing each of those lines
to a single member of the range. Finally, in
Example 2.9, the domain breaks into
planes which get "zeroed out", and so the map starts with a
three-dimensional domain but ends with a one-dimensional range—
this map "subtracts" two from the dimension. (Notice that, in this
third example, the codomain is two-dimensional but the range of the
map is only one-dimensional, and it is the dimension of the range that
is of interest.)
- Theorem 2.14
A linear map's rank plus its nullity equals the dimension of its domain.
Let be linear and let be a basis for the nullspace. Extend that to a basis for the entire domain. We shall show that is a basis for the rangespace. Then counting the size of these bases gives the result.
To see that is linearly independent, consider the equation . This gives that and so is in the nullspace of . As is a basis for this nullspace, there are scalars satisfying this relationship.
But is a basis for so each scalar equals zero. Therefore is linearly independent.
To show that spans the rangespace, consider and write as a linear combination of members of . This gives and since , ..., are in the nullspace, we have that . Thus, is a linear combination of members of , and so spans the space.
- Example 2.15
the rangespace and nullspace are
and so the rank of is two while the nullity is one.
- Example 2.16
If is the linear transformation then the range is , and so the rank of is one and the nullity is zero.
- Corollary 2.17
The rank of a linear map is less than or equal to the dimension of the domain. Equality holds if and only if the nullity of the map is zero.
that an isomorphism exists between two spaces
if and only if their dimensions are equal.
Here we see that for a homomorphism to exist,
the dimension of the range must be less than or equal to the
dimension of the domain.
For instance, there is no homomorphism
from onto .
There are many homomorphisms
from into , but none is onto all
The rangespace of a linear map can be of dimension strictly less than
the dimension of the domain
(Example 2.3's derivative transformation on
has a domain of dimension four but a range of dimension three).
Thus, under a homomorphism,
linearly independent sets in the domain
may map to linearly dependent sets in the range
(for instance, the derivative sends
That is, under a homomorphism, independence may be lost.
In contrast, dependence stays.
- Lemma 2.18
Under a linear map, the image of a linearly dependent set is linearly dependent.
Suppose that , with some nonzero. Then, because and because , we have that with some nonzero .
When is independence not lost?
One obvious sufficient condition is when the homomorphism is an isomorphism.
This condition is also necessary;
see Problem 14.
We will finish this subsection comparing homomorphisms with isomorphisms
by observing that
a one-to-one homomorphism is an isomorphism from its domain onto its range.
- Definition 2.19
A linear map that is one-to-one is nonsingular.
(In the next section we will see the connection
between this use of "nonsingular" for maps and its familiar
use for matrices.)
- Example 2.20
This nonsingular homomorphism
gives the obvious correspondence between and the -plane inside of .
The prior observation allows us to adapt some results about isomorphisms
to this setting.
- Theorem 2.21
In an -dimensional vector space , these:
- is nonsingular, that is, one-to-one
- has a linear inverse
- , that is,
- if is a basis for then is a basis for
statements about a linear map .
We will first show that . We will then show that .
For , suppose that the linear map is one-to-one, and so has an inverse. The domain of that inverse is the range of and so a linear combination of two members of that domain has the form . On that combination, the inverse gives this.
Thus the inverse of a one-to-one linear map is automatically linear. But this also gives the implication, because the inverse itself must be one-to-one.
Of the remaining implications, holds because any homomorphism maps to , but a one-to-one map sends at most one member of to .
Next, is true since rank plus nullity equals the dimension of the domain.
For , to show that is a basis for the rangespace we need only show that it is a spanning set, because by assumption the range has dimension . Consider . Expressing as a linear combination of basis elements produces , which gives that , as desired.
Finally, for the implication, assume that is a basis for so that is a basis for . Then every a the unique representation . Define a map from to by
(uniqueness of the representation makes this well-defined). Checking that it is linear and that it is the inverse of are easy.
We've now seen that a linear map
shows how the structure of the domain is like that of the range.
Such a map can be thought to organize the domain space into
inverse images of points in the range.
In the special case that the map is one-to-one, each inverse image is a single
point and the map is an isomorphism between the domain and the range.