Linear Algebra/Rangespace and Nullspace

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 $$ \iota:\mathbb{R}^2\to \mathbb{R}^3 $$



\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} $$

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 $$\iota$$ is onto the $$xy$$-plane subset of $$ \mathbb{R}^3$$.

(We shall soon see the connection between the rank of a map and the rank of a matrix.)

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 $$h:V\to W$$, the set of elements of $$V$$ that are mapped to $$ \vec{w}\in W $$ is the inverse image $$h^{-1}(\vec{w})=\{\vec{v}\in V\,\big|\, h(\vec{v})=\vec{w}\}$$. Above, the three sets of many elements on the left are inverse images.

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 $$\vec{w}$$. 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.

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, and 2.9.

First, in all three examples, the inverse images are lines or planes, that is, linear surfaces. In particular, the inverse image of the range's zero vector is a line or plane through the origin— a subspace of the domain.



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— $$\pi$$, 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.)

We know 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 $$ \mathbb{R}^2 $$ onto $$ \mathbb{R}^3 $$. There are many homomorphisms from $$ \mathbb{R}^2 $$ into $$ \mathbb{R}^3 $$, but none is onto all of three-space.

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 $$\mathcal{P}_3$$ 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 $$\{1,x,x^2,x^3\}$$ to $$\{0,1,2x,3x^2\}$$). That is, under a homomorphism, independence may be lost. In contrast, dependence stays.

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.

(In the next section we will see the connection between this use of "nonsingular" for maps and its familiar use for matrices.)

The prior observation allows us to adapt some results about isomorphisms to this setting.

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.

Exercises
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