We start with two examples that suggest the right definition.
Consider the example mentioned above, the space of two-wide row
vectors and the space of two-tall column vectors. They are "the same"
in that if we associate the vectors that have the same components,
then this correspondence preserves the operations, for instance this addition
and this scalar multiplication.
More generally stated, under the correspondence
both operations are preserved:
(all of the variables are real numbers).
Another two spaces we can think of as "the same" are
, the space of quadratic polynomials,
and . A natural correspondence is this.
The structure is preserved: corresponding elements add in a corresponding way
and scalar multiplication corresponds also.
An isomorphism between two vector spaces and is a map that
(we write , read " is isomorphic to ", when such a map exists).
("Morphism" means map, so "isomorphism" means a map expressing sameness.)
The vector space of functions of is isomorphic to the vector space under this map.
We will check this by going through the conditions in the definition.
We will first verify condition 1, that the map is a correspondence between the sets underlying the spaces.
To establish that is one-to-one, we must prove that only when . If
then, by the definition of ,
from which we can conclude that and because column vectors are equal only when they have equal components. We've proved that implies that , which shows that is one-to-one.
To check that is onto we must check that any member of the codomain is the image of some member of the domain . But that's clear—any
is the image under of .
Next we will verify condition (2), that preserves structure.
This computation shows that preserves addition.
A similar computation shows that preserves scalar multiplication.
With that, conditions (1) and (2) are verified, so we know that is an isomorphism and we can say that the spaces are isomorphic .
Let be the space of linear combinations of three variables , , and , under the natural addition and scalar multiplication operations. Then is isomorphic to , the space of quadratic polynomials.
To show this we will produce an isomorphism map. There is more than one possibility; for instance, here are four.
The first map is the more natural correspondence in that it just carries the coefficients over. However, below we shall verify that the second one is an isomorphism, to underline that there are isomorphisms other than just the obvious one (showing that is an isomorphism is Problem 3).
To show that is one-to-one, we will prove that if then . The assumption that gives, by the definition of , that .
Equal polynomials have equal coefficients, so , , and . Thus implies that and therefore is one-to-one.
The map is onto because any member of the codomain is the image of some member of the domain, namely it is the image of . For instance, is .
The computations for structure preservation are like those in the prior example. This map preserves addition
and scalar multiplication.
Thus is an isomorphism and we write .
We are sometimes interested in an isomorphism
of a space with itself, called an automorphism. An identity map
is an automorphism. The next two examples show that there are others.
A dilation map that multiplies all vectors by a nonzero scalar is an automorphism of .
A rotation or turning map that rotates all vectors through an angle is an automorphism.
A third type of automorphism of is a map that flips or reflects all vectors over a line through the origin.
Consider the space of polynomials of degree 5 or less and the map that sends a polynomial to . For instance, under this map and . This map is an automorphism of this space; the check is Problem 12.
This isomorphism of with itself does more than just tell us that the space is "the same" as itself. It gives us some insight into the space's structure. For instance, below is shown a family of parabolas, graphs of members of . Each has a vertex at , and the left-most one has zeroes at and , the next one has zeroes at and , etc.
Geometrically, the substitution of for in any function's argument shifts its graph to the right by one. Thus, and 's action is to shift all of the parabolas to the right by one. Notice that the picture before is applied is the same as the picture after is applied, because while each parabola moves to the right, another one comes in from the left to take its place. This also holds true for cubics, etc. So the automorphism gives us the insight that has a certain horizontal homogeneity; this space looks the same near as near .
As described in the preamble to this section, we will next
produce some results supporting the
contention that the definition of isomorphism above
captures our intuition of vector spaces being the same.
Of course the definition itself is persuasive: a vector space
consists of two components, a set and some structure,
and the definition simply requires
that the sets correspond and that the structures correspond also.
Also persuasive are the examples above.
In particular, Example 1.1, which
gives an isomorphism between the space of two-wide row vectors and
the space of two-tall column vectors,
dramatizes our intuition that isomorphic spaces are
the same in all relevant respects.
Sometimes people say, where , that " is just
painted green"—any differences are merely cosmetic.
Further support for the definition, in case it is needed,
is provided by the following results that, taken together, suggest that
all the things of interest in a vector space
correspond under an isomorphism.
Since we studied vector spaces to study linear combinations,
"of interest" means "pertaining to linear combinations".
Not of interest is the way that the vectors are presented
typographically (or their color!).
As an example, although the definition of isomorphism doesn't explicitly
say that the zero vectors must correspond,
it is a consequence of that definition.
An isomorphism maps a zero vector to a zero vector.
Where is an isomorphism, fix any .
The definition of isomorphism requires that sums of two vectors correspond and
that so do scalar multiples.
We can extend that to say that all linear combinations correspond.
For any map between vector spaces these statements are equivalent.
preserves linear combinations of two vectors
preserves linear combinations of any finite number of vectors
Since the implications and are clear, we need only show that . Assume statement 1. We will prove statement 3 by induction on the number of summands .
The one-summand base case, that , is covered by the assumption of statement 1.
For the inductive step assume that statement 3 holds whenever there are or fewer summands, that is, whenever , or , ..., or . Consider the -summand case. The first half of 1 gives
by breaking the sum along the final "". Then the inductive hypothesis lets us break up the -term sum.
Finally, the second half of statement 1 gives
when applied times.
In addition to adding to the intuition that the definition of isomorphism
does indeed preserve the things of interest in a vector space,
that lemma's second item is an especially handy way of
checking that a map preserves structure.
We close with a summary.
The material in this section augments the chapter on Vector Spaces.
There, after giving the definition of a vector space,
we informally looked at what different things can happen.
Here, we defined the relation
"" between vector spaces and
we have argued that it
is the right way to split the collection of vector spaces
into cases because it preserves the features of interest in a vector
space—in particular, it preserves linear combinations.
That is, we have now said precisely what we mean by "the same",
and by "different", and so we have precisely classified the vector spaces.
We show that isomorphisms can be tailored to fit in that, sometimes, given vectors in the domain and in the range we can produce an isomorphism associating those vectors.
Let be a basis for so that any has a unique representation as , which we denote in this way.
Show that the operation is a function from to (this entails showing that with every domain vector there is an associated image vector in , and further, that with every domain vector there is at most one associated image vector).
Show that this function is one-to-one and onto.
Show that it preserves structure.
Produce an isomorphism from to
that fits these specifications.
Prove that a space is -dimensional if and only if it is isomorphic to . Hint. Fix a basis for the space and consider the map sending a vector over to its representation with respect to .
(Requires the subsection on Combining Subspaces, which is optional.) Let and be vector spaces. Define a new vector space, consisting of the set along with these operations.
This is a vector space, the external direct sum of and .
Check that it is a vector space.
Find a basis for, and the dimension of, the external direct sum .
What is the relationship among , , and ?
Suppose that and are subspaces of a vector space such that (in this case we say that is the internal direct sum of and ). Show that the map given by
is an isomorphism. Thus if the internal direct sum is defined then the internal and external direct sums are isomorphic.