Serlo: EN: Isomorphisms

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Isomorphic Structures and Isomorphisms

Isomorphic Structures

We consider the vector space $K[X{]}_{\le 2}$ of polynomials of degree less than or equal to $2$ and we consider ${ℝ}^3$. Vectors in these spaces have a one-to-one correspondence, as we have already seen in the introduction article to vector spaces: {{#lst:Serlo: EN:_Introduction: Vector space|polynom_vektor}} We also found that addition and scalar multiplication work the same way in both vector spaces: {{#lst:Serlo: EN:_Introduction: Vector space|polynom-vektor_1}} {{#lst:Serlo: EN:_Introduction: Vector space|polynom-vektor_2}}

In general, vector spaces can be thought of as sets with some structure. In our example, we can match the sets 1 to 1. And also the structures (i.e. addition and multiplication) can be matched. So both vector spaces "essentially carry the same information", although they formally comprise different objects. In such a case, we will call the two vector spaces isomorphic (to each other). The bijection which identifies the two vector spaces is then called an isomorphism.

We now derive what the mathematical definition is of "two vector spaces $V$ and $W$ are isomorphic":

The identification of the ``sets`` is given by a bijective mapping $f:V\to W$. Preserving the structure means that addition and scalar multiplication are preserved when mapping back and forth with $f$ and $f^{-1}$. But "preserving addition and scalar multiplication" for a mapping between vector spaces is nothing else than "being linear". So we want $f$ and $f^{-1}$ to be linear.

Let us now return to our example from above. In this case, the identification map we are looking for from the Definition would look like this:

Isomorphism

We also want to give a name to the map $f$ introduced above:

Alternative Derivation Vorlage:Anker

Now let's look at the term "vector space" from a different point of view. We can also think of a vector space as a basis together with corresponding linear combinations of the basis. So we can call vector spaces "equal" if we can identify the bases 1 to 1 and the corresponding linear combinations are generated in the same way. In other words, we are looking for a mapping that preserves both bases and linear combinations. What property must the mapping have in order to generate the same linear combinations? The answer is almost in the name: The mapping must be linear.

Let us now turn to the question of what property a linear map needs in order to map bases to bases. A basis is nothing else than a linearly independent generator. Thus, the map must preserve generators and linear independence. A linear map that preserves a generator is called an epimorphism - that is, a surjective linear map. A linear map that preserves linear independence is called a monomorphism and is thus an injective linear map. So the function we are looking for is an epimorphism and a monomorphism at the same time. As a monomorphism it must be injective. As an epimorphism, on the other hand, the mapping must be surjective. So overall we get a bijective linear map. This we again call an isomorphism. This gives us the alternative definition:

Inverse Mappings of Linear Bijections are Linear

We have derived two descriptions for isomorphisms. Thus we have also two different definitions. The first one seems to require more than the second one: In the first definition, an isomorphism $f$ must additionally satisfy that $f^{-1}$ is linear. Does this give us two different mathematical objects, or does linearity of $f$ already imply linearity of $f^{-1}$? According to our intuition, both definitions should define the same objects. So $f$ being linear should then imply $f^{-1}$ being linear. And indeed, this is the case:

Classifying Isomorphic Structures

Bijections of Bases Generate an Isomorphism

In the alternative derivation , we used the intuition that an isomorphism is a linear map that "preserves bases". This means that bases are sent to bases and linear combinations are preserved. So, describing it a bit more formally, we considered the following:

We already know the following: If $f:V\to W$ is a linear map between two vector spaces and $f$ is an isomorphism, then $f$ maps bases of $V$ to bases of $W$.

But we don't know yet whether a linear map that sends a basis to a basis, is already an isomorphism. This statement indeed turns out to be true.

If we have given a bijection between bases, then there is a nice description of the inverse of $f$: We know that $f^{-1}$ is characterized by the conditions $f^{-1}\circ f={\mathrm{id}}_V$ and $f\circ f^{-1}={\mathrm{id}}_W$. Further, the principle of linear continuation tells us that we need to know $f^{-1}$ only on a basis of $W$ to describe it completely. Now we have already chosen the basis $B_W$ of W. That is, we are interested in $f^{-1}(b_W)$ for $b_W\in B_W$. Because $h$ is bijective, there is exactly one $b_V\in B_V$ with $h(b_V)=b_W$. Therefore, we get $f^{-1}(b_W)=f^{-1}(f(b_V))=b_V$ from the above conditions. Now how can we describe this element $b_V$ more precisely? $b_V$ is the unique preimage of $b_W$ under $h$. So $b_V=h^{-1}(b_W)$. In other words, $f^{-1}$ is the linear map induced by $h^{-1}$ from $W$ to $V$.

Classification of Finite Dimensional Vector Spaces

When are two finite-dimensional vector spaces isomorphic? If $V$ and $W$ are finite-dimensional vector spaces, then we have bases $\{b_1,\dots ,b_n\}$ of $V$ and $\{c_1,\dots ,c_m\}$ of $W$. From the previous theorem we know that an isomorphism is uniquely characterized by the bijection of the bases. When do we find a bijection between these two sets? Exactly when they have the same size, so $n=m$. Or in other words, if $V$ and $W$ have the same dimension:

We have shown that all $K$-vector spaces of dimension $n$ are isomorphic. In particular, all such vector spaces are isomorphic to the vector space $K^n$. Because the $K^n$ is a well-describable model for a vector space, let us examine in more detail the isomorphism constructed in the last theorem.

Let $V$ be an $n$-dimensional $K$-vector space. We now follow the proof of the last theorem to understand the construction of the isomorphism. We use that bases of $V$ and of $K^n$ have the same size. For the isomorphism, we construct a bijection between a basis of $V$ and a basis of $K^n$. The space $K^n$ has as kind of "standard basis", given by the canonical basis $\{e_1,\dots ,e_n\}$.

Following the proof of the last theorem, we see that we must choose a basis of $V$ and a basis of $K^n$. For $K^n$ we choose the standard basis $E$ and for $V$ we choose some basis $B$ of $V$. Next, we need a bijection $h:E\to B$ between the standard basis and the basis $B$. That is, we need to associate exactly one $b\in B$ with each $e_i$. We can thus name the images of $e_i$ as $b_i=h(e_i)$. Because $h$ is bijective, we get $B=\{b_1,\dots ,b_n\}$. In essence, we have used this to number the elements of B. Mathematically, numbering the elements of $B$ is the same as giving a bijection from $E$ to $B$, since we can simply map $e_i$ to the $i$-th element of $B$.

The principle of linear continuation now provides us with an isomorphism $f:K^n\to V$. By linear continuation, this isomorphism sends the vector $(x_1,\dots ,x_n{)}^T\in K^n$ to the element $f((x_1,\dots ,x_n{)}^T)=x_1\cdot b_1+\dots +x_n\cdot b_n$.

Now what about the map that sends $B$ to $E$, i.e., the inverse map $f^{-1}$ of $f$?

We have already computed above what the mapping $f^{-1}$ looks like in this case. $f^{-1}$ is just the mapping induced by $h^{-1}$ via the principle of linear continuation. That is, for basis vectors, we know that $f$ maps $b_i\in B$ to $h^{-1}(b_i)=e_i$. And where does it map a general vector $v\in V$? HEre, we use the principle of linear continuation: We write $v$ as a linear combination of our basis $v={\lambda }_1{b}_1+\dots +{\lambda }_n{b}_n$. By linearity, the mapping $f^{-1}$ now sends $v$ to $f^{-1}(v)=({\lambda }_1,\dots ,{\lambda }_n{)}^T$. In particular, the ${\lambda }_i$ describe where $v$ is located with respect to the basis vectors $b_i$. This is just like GPS coordinates, which tells you your position with respect to certain anchor points (there prime meridian and equator). Therefore, we can say that $f^{-1}$ sends each vector to its coordinates with respect to the basis $B$.

We now want to investigate how many choices the construction of the coordinate map depends on.

Mathe für Nicht-Freaks: Vorlage:Beispiel The coordinate mapping depends on the choice of the basis. If you have different bases, you get different mappings.

Mathe für Nicht-Freaks: Vorlage:Beispiel Even if we only change the numbering of the elements of a base, we already get different coordinate mappings. Mathe für Nicht-Freaks: Vorlage:Beispiel

In order to speak of the coordinate mapping, we must also specify the order of the basis elements. A basis where we also specify the order of the basis elements is called an ordered basis. Mathe für Nicht-Freaks: Vorlage:Definition With this notion we can simplify the notation of the coordinate mapping. If $B=(b_1,\dots ,b_n)$ is an ordered basis, we also denote the coordinate mapping $k_{b_1,\dots ,b_n}$ as $k_B$.

We have now talked about a class of isomorphisms from $V$ to $K^n$. Are there any other isomorphisms from $V$ to $K^n$? That is, are there isomorphisms that are not coordinate mappings? In fact, every isomorphism from $V$ to $K^n$ is a coordinate mapping with respect to a proper basis.

Exercises

<section begin=aufgaben_isomorphismus /> Mathe für Nicht-Freaks: Vorlage:Aufgabe

Mathe für Nicht-Freaks: Vorlage:Aufgabe

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