Isomorphisms – Serlo

Isomorphic Structures and Isomorphisms

Isomorphic Structures

We consider the vector space $K[X]_{\leq 2}$ of polynomials of degree less than or equal to $2$ and we consider $\mathbb {R} ^{3}$. Vectors in these spaces have a one-to-one correspondence, as we have already seen in the introduction article to vector spaces:

${\begin{array}{cccc}f(x)&=&{\color {Red}7}x^{2}&+{\color {OliveGreen}3}x&+({\color {NavyBlue}-2})&\leftrightarrow &({\color {Red}7},{\color {OliveGreen}3},{\color {NavyBlue}-2})^{T}\\g(x)&=&{\color {Red}1}x^{2}&+{\color {OliveGreen}0}x&+{\color {NavyBlue}6}&\leftrightarrow &({\color {Red}1},{\color {OliveGreen}0},{\color {NavyBlue}6})^{T}\end{array}}$

We also found that addition and scalar multiplication work the same way in both vector spaces:

${\begin{array}{ccc}{\color {Red}7}x^{2}&+{\color {OliveGreen}3}x&+({\color {NavyBlue}-2})&\leftrightarrow &({\color {Red}7},{\color {OliveGreen}3},{\color {NavyBlue}-2})^{T}\\&+&&&+\\{\color {Red}1}x^{2}&+{\color {OliveGreen}0}x&+{\color {NavyBlue}6}&\leftrightarrow &({\color {Red}1},{\color {OliveGreen}0},{\color {NavyBlue}6})^{T}\\&=&&&=\\{\color {Red}8}x^{2}&+{\color {OliveGreen}3}x&+{\color {NavyBlue}4}&\leftrightarrow &({\color {Red}8},{\color {OliveGreen}3},{\color {NavyBlue}4})^{T}\end{array}}$

${\begin{array}{ccc}&-1&&&-1\\&\cdot &&&\cdot \\{\color {Red}7}x^{2}&+{\color {OliveGreen}3}x&+({\color {NavyBlue}-2})&\leftrightarrow &({\color {Red}7},{\color {OliveGreen}3},{\color {NavyBlue}-2})^{T}\\&=&&&=\\{\color {Red}-7}x^{2}&+({\color {OliveGreen}-3})x&+{\color {NavyBlue}2}&\leftrightarrow &({\color {Red}-7},{\color {OliveGreen}-3},{\color {NavyBlue}2})^{T}\\\end{array}}$

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.

Definition (Isomorphic)

The vector spaces $V$ and $W$ are isomorphic if there is a bijective map $f\colon V\to W$ between them such that $f$ and $f^{-1}$ are linear. We then write $V\cong W$.

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:

${\begin{aligned}f\colon \mathbb {R} ^{3}&\to K[X]_{\leq 2}\\[0.3em](a,b,c)&\mapsto ax^{2}+bx+c\end{aligned}}$

Isomorphism

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

Definition (Isomorphism)

An isomorphism between vector spaces $V$ and $W$ is a bijective map $f\colon V\to W$ such that $f$ and $f^{-1}$ are linear.

Alternative Derivation

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:

Definition (Alternative definition of isomorphism and isomorphism)

Two vector spaces $V$ and $W$ are isomorphic if there is a bijective linear map $f\colon V\to W$ between them.

A map $f\colon V\to W$ is called an isomorphism if it is a bijective linear map.

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:

Theorem (The inverse map of a bijective linear map is again linear.)

Let $f\colon V\to W$ be a bijective linear map. Then the inverse mapping $g\colon W\to V$ is also linear.

How to get to the proof? (The inverse map of a bijective linear map is again linear.)

We want to show that $g$ is linear. For this, both $g(w+w')=g(w)+g(w')$ and $g(\lambda \cdot w)=\lambda \cdot g(w)$ must hold for all vectors $w,w'\in W$ and scalars $\lambda \in K$.

We have given that $f$ is linear and bijective with inverse map $g$. How can we use this to show the linearity of $g$? Since $g$ is the inverse map of $f$, we have:

${\begin{aligned}&f\circ g=id_{W}{\text{, so }}f(g(w))=w{\text{ for all }}w\in W\end{aligned}}$

Together with the linearity of $f$, this gives us:

${\begin{aligned}g(w+w')&=g(f(g(w))+f(g(w')))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{ linearity of }}f\right.}\\[0.3em]&=g(f(g(w)+g(w')))=g(w)+g(w').\end{aligned}}$

In the same way we can proceed for homogeneity $g(\lambda \cdot w)=\lambda \cdot g(w)$.

Proof (The inverse map of a bijective linear map is again linear.)

For the inverse $g$ of $f$ , it holds that:

${\begin{aligned}&f\circ g=id_{W};\\[0.3em]&g\circ f=id_{V}.\end{aligned}}$

So for every vector $v\in V$ and every vector $w\in W$

${\begin{aligned}&(f\circ g)(w)=f(g(w))=w;\\[0.3em]&(g\circ f)(v)=g(f(v))=v.\end{aligned}}$

Proof step: $g$ is additive.

Let $w$ and $w'\in W$ be two vectors. Then we have:

${\begin{aligned}&g(w+w')\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \operatorname {g} {\text{ is the inverse of }}\operatorname {f} \right.}\\[0.3em]=\,&g(f(g(w))+f(g(w')))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \operatorname {f} {\text{ is linear }}\right.}\\[0.3em]=\,&g(f(g(w)+g(w'))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \operatorname {g} {\text{ is the inverse of }}\operatorname {f} \right.}\\[0.3em]=\,&g(w)+g(w').\end{aligned}}$

Thus the inverse function is additive.

Proof step: $g$ Is homogeneous.

Let $w\in W$ be a vector and $\lambda \in K$ is a scalar. Then we have that

${\begin{aligned}&g(\lambda \cdot w)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \operatorname {g} {\text{ is the inverse of }}\operatorname {f} \right.}\\[0.3em]=\,&g(\lambda \cdot f(g(w)))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \operatorname {f} {\text{ is linear }}\right.}\\[0.3em]=\,&g(f(\lambda \cdot g(w)))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \operatorname {g} {\text{ is the inverse of }}\operatorname {f} \right.}\\[0.3em]=\,&\lambda \cdot g(w).\end{aligned}}$

Thus the inverse function is homogeneous.

Thus have shown that the inverse $g$ is also linear.

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 $f\colon V\to W$ 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.

Theorem

Let $K$ be a field, $V,W$ two $K$-vector spaces, $B_{V}$ a basis of $V$ and $f:V\to W$ a linear map.

Then $f$ is an isomorphism if and only if $B_{V}$ is mapped by $f$ to a basis of $W$.

Proof

Proof step: $\implies$

Let $f$ be an isomorphism. Then $f$ is by definition both a monomorphism and an epimorphism.

We want to show that $f$ preserves bases. That is, the image of $B_{V}$ under $f$ is a linearly independent generator of $W$.

Proof step: $f(B_{V})$ is linearly independent

We know from the article on monomorphisms that those preserve linear independence. The set $B_{V}$ is a basis and thus linearly independent. So its image under $f$ is also linearly independent.

Proof step: $f(B_{V})$ is a generator of $W$

We know from the article on epimorphisms that those preserve generators . The set $B_{V}$ is a basis and hence a generator of $V$. So its image under $f$ is a generator of $W$.

Proof step: $\Longleftarrow$

$f$ maps $B_{V}$ to a basis $B_{W}$ of $W$ .

Proof step: Injectivity

Since $f$ maps the linearly independent set $B_{V}$ to the linearly independent set $B_{W}$, $f$ preserves linear independence. From the article on monomorphisms we know that $f$ must thus be injective.

Proof step: Surjectivity

$f$ maps the basis $B_{V}$, (which is in particular a generator), to the basis $B_{W}$ (which is also a generator). From the article on epimorphisms we know that $f$ must thus be surjective.

$f$ is linear by premise. Together with injectivity and surjectivity it follows that $f$ is an isomorphism.

Theorem

Let $V$ and $W$ be two $K$-vector spaces with bases$B_{V}$ and $B_{W}$. Let further $h\colon B_{V}\to B_{W}$ be a bijective mapping. Then there is exactly one isomorphism $f\colon V\to W$ with $f|_{B_{V}}=h$.

Proof

From the article about linear continuation we know that we can find a unique linear map $f$ with $f(b)=h(b)\in B_{W}$ for all $b\in B_{V}$. Thus, as required by the premise, $f|_{B_{V}}=h$.

We still have to show that the mapping $f$ is an isomorphism. By the previous theorem, we must show that $f$ maps a basis of $V$ to a basis of $W$. Now we have constructed $f$ exactly such that $f|_{B_{V}}=h$. That is, $f$ maps the basis $B_{V}$ to the basis $B_{W}$ since $h$ is bijective. So $f$ is an isomorphism.

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 $f^{-1}\circ f=\operatorname {id} _{V}$ and $f\circ f^{-1}=\operatorname {id} _{W}$. 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 $\{b_{1},\ldots ,b_{n}\}$ of $V$ and $\{c_{1},\ldots ,c_{m}\}$ 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:

Theorem (Finite dimensional vector spaces with the same dimension are isomorphic)

Let $V,W$ be finite-dimensional vector spaces. Then: $\operatorname {dim} (V)=\operatorname {dim} (W)\Longleftrightarrow V\cong W$

Proof (Finite dimensional vector spaces with the same dimension are isomorphic)

Proof step: $\implies$

Let $\operatorname {dim} (V)=\operatorname {dim} (W)$.

Two vector spaces are called isomorphic if there exists an isomorphism between them. We know that an isomorphism exists between vector spaces if we can find a bijective mapping between the bases of them. Since $\operatorname {dim} (V)=\operatorname {dim} (W)$, we find a bijective mapping between bases. Thus, there exists an isomorphism between $V$ and $W$.

Thus, $V$ and $W$ are isomorphic.

Proof step: $\Longleftarrow$

Let $V\cong W$.

Let $f$ be an isomorphism between $V$ and $W$. We know that an isomorphism maps bases to bases. That is, $f(B_{V})$ is a basis of $W$. In particular, since the mapping is an isomorphism, it is bijective. Thus $|B_{V}|=|f(B_{V})|$.

This implies $\operatorname {dim} (V)=\operatorname {dim} (W)$.

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 $h\colon E\to B$ 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 $f\colon K^{n}\to V$. 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 $B$.

Definition (Coordinate mapping)

Let $V$ be a $n$-dimensional $K$-vector space and $b_{1},\dots ,b_{n}$ a basis of $V$. We define the isomorphism $k_{b_{1},\dots ,b_{n}}\colon V\to K^{n}$ as the continuation of the following bijection between the base $b_{1},\dots ,b_{n}$ and the standard basis of $K^{n}$:

$\{b_{1},\dots ,b_{n}\}\to \{e_{1},\dots ,e_{n}\};b_{i}\mapsto e_{i}$

We call $k_{b_{1},\dots ,b_{n}}$ the coordinate mapping with respect to $b_{1},\dots ,b_{n}$.

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

Example (Coordinate mapping between the vector space of real quadratic polynomials and $\mathbb {R} ^{3}$)

We consider the two $\mathbb {R}$-vector spaces $W:=\mathbb {R} ^{3}$ and that of real polynomials of degree $\leq 2:=P_{\leq 2}$. The coordinate mapping then looks like this:

$k_{1,x,x^{2}}:P_{2}\to W$, $a_{0}+a_{1}x+a_{2}x^{2}\mapsto (a_{0},a_{1},a_{2})$.

The coordinate mapping depends on the choice of the basis. If you have different bases, you get different mappings.

Example (Different bases create different coordinate mappings)

We consider the following two bases of $\mathbb {R} ^{2}$: $B=\{(1,0)^{T},(0,1)^{T}\}$ and $C=\{(1,1)^{T},(1,-1)^{T}\}$.

For $x=(x_{1},x_{2})\in \mathbb {R} ^{2}$ we have

$x=x_{1}\cdot {\begin{pmatrix}1\\0\end{pmatrix}}+x_{2}\cdot {\begin{pmatrix}0\\1\end{pmatrix}}$

So the coordinate mapping with respect to $B$ looks like this

$k_{(1,0)^{T},(0,1)^{T}}\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\mapsto {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}$

For the base $C$ we have

$x={\frac {x_{1}+x_{2}}{2}}\cdot {\begin{pmatrix}1\\1\end{pmatrix}}+{\frac {x_{1}-x_{2}}{2}}\cdot {\begin{pmatrix}1\\-1\end{pmatrix}}$

Thus, the coordinate mapping with respect to $C$ is.

$k_{(1,1)^{T},(1,-1)^{T}}\colon {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\mapsto {\begin{pmatrix}{\frac {x_{1}+x_{2}}{2}}\\{\frac {x_{1}-x_{2}}{2}}\end{pmatrix}}.$

These two mappings are not the same. For example

$k_{(1,0)^{T},(0,1)^{T}}\left({\begin{pmatrix}1\\3\end{pmatrix}}\right)={\begin{pmatrix}1\\3\end{pmatrix}}\neq {\begin{pmatrix}2\\-1\end{pmatrix}}=k_{(1,1)^{T},(1,-1)^{T}}\left({\begin{pmatrix}1\\3\end{pmatrix}}\right)$

Even if we only change the numbering of the elements of a base, we already get different coordinate mappings.

Example (Different numbering of the basis result in different coordinate images)

We consider the standard basis $B=\{(1,0)^{T},(0,1)^{T}\}$ of $\mathbb {R} ^{2}$. We want to find out what the coordinate mappings $k_{(1,0)^{T},(0,1)^{T}}$ and $k_{(0,1)^{T},(1,0)^{T}}$ look like. For $k_{(1,0)^{T},(0,1)^{T}}$ we already know this:

$k_{(1,0)^{T},(0,1)^{T}}\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\mapsto {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}$

For $x=(x_{1},x_{2})^{T}$ we have

$x=x_{2}\cdot {\begin{pmatrix}0\\1\end{pmatrix}}+x_{1}\cdot {\begin{pmatrix}1\\0\end{pmatrix}}.$

The construction of the coordinate mapping thus provides us with the following description

$k_{(0,1)^{T},(1,0)^{T}}\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\mapsto {\begin{pmatrix}x_{2}\\x_{1}\end{pmatrix}}$

These two mappings are different. For example

$k_{(1,0)^{T},(0,1)^{T}}\left({\begin{pmatrix}2\\1\end{pmatrix}}\right)={\begin{pmatrix}2\\1\end{pmatrix}}\neq {\begin{pmatrix}1\\2\end{pmatrix}}=k_{(0,1)^{T},(1,0)^{T}}\left({\begin{pmatrix}2\\1\end{pmatrix}}\right).$

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.

Definition (Ordered basis)

Let $K$ be a field and $V$ a finite-dimensional $K$-vector space. Let $B=(b_{1},\dots ,b_{n})\in V^{n}$. Then we call $B$ an ordered basis of $V$ if $\{b_{1},\dots ,b_{n}\}$ is a basis of $V$.

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.

Theorem (All isomorphisms $V\to K^{n}$ are coordinate mappings)

Let $f\colon V\to K^{n}$ be an isomorphism. Then there is exactly one ordered basis $B$ of $V$ such that $f=k_{B}$.

How to get to the proof? (All isomorphisms $V\to K^{n}$ are coordinate mappings)

We have constructed the coordinate mapping as an inverse mapping. For this we bijectively mapped the standard basis of $K^{n}$ to a basis of $V$. To reconstruct this basis, we need to consider the preimages of the standard basis under $f$. That is, we need $B=f^{-1}(E)$, which requires choosing an ordering of $B$. For instance, we may set $b_{i}=f^{-1}(e_{i})$. Now, we know that $B$ is a basis because $f^{-1}$ is an isomorphism. Further, we have just above applied the principle of linear continuation backwards, which told us that all of $f^{-1}$ is induced by only the bijection $e_{i}\mapsto b_{i}$. Further above, we have also seen that $f$ is already induced by the bijection $b_{i}\mapsto e_{i}$. But this gives exactly the coordinate mapping with respect to $B$.

Proof (All isomorphisms $V\to K^{n}$ are coordinate mappings)

We define $b_{i}:=f^{-1}(e_{i})$ for $1\leq i\leq n$. Then $\{b_{1},\dots ,b_{n}\}$ is the image of the standard basis under the mapping $f^{-1}$. Since $f^{-1}$ is an isomorphism, it maps bases to bases. Thus $b_{1},\ldots ,b_{n}$ is a basis of $V$.

Define the ordered basis $B:=(b_{1},\dots ,b_{n})$. We now show $f=k_{B}$. For this it is sufficient to prove equality on the basis $B$, since $f$ and $k_{B}$ are linear. For any $b_{i}$ it holds that

$f(b_{i})=f(f^{-1}(e_{i}))=e_{i}=k_{B}(b_{i}).$

So indeed, $f=k_{B}$.

Examples of vector space isomorphisms

Example (Real polynomials of $n$-th degree and $\mathbb {R} ^{n+1}$)

For $n=2$ , we can establish an isomorphism between the space of polynomials of at most second degree $\mathbb {R} [X]_{\leq 2}$ and the space $\mathbb {R} ^{3}$.

We define the mapping $f:\mathbb {R} [X]_{\leq 2}\to \mathbb {R} ^{3}$ vis $f(p)=(p(-1),\,p(0),\,p(1))^{T}$.

Claim: $f$ is an isomorphism.

For this, we need to prove three things:

1. $f$ is a linear map $f:\mathbb {R} [X]_{\leq 2}\to \mathbb {R} ^{3}$
2. $f$ is injektive
3. $f$ is surjektive

Proof step: Linearity of $f$

Since $f(p)$ is defined for every polynomial $p\in \mathbb {R} [X]_{\leq 2}$ and has values in $\mathbb {R} ^{3}$, $f$ is well-defined as a mapping.

So we still have to prove that for $p_{1},p_{2}\in \mathbb {R} [X]_{\leq 2}$ and $\lambda \in \mathbb {R}$ it always holds that $f(p_{1}+p_{2})=f(p_{1})+f(p_{2})$ and $f(\lambda p)=\lambda f(p)$.

This is completely analogous to this calculation.

Proof step: Injectivity of $f$

Let $p\in \mathbb {R} [X]_{\leq 2}$ and $f(p)=(0,0,0)^{T}$.

This means that the polynomial of the highest second degree $p$ has three zeros:$p(-1)=p(0)=p(1)=0$. It follows (e.g., with polynomial division) that we can write $p$ as $p(x)=(x+1)\cdot (x-0)\cdot (x-1)\cdot q(x)$, where $q$ is again a polynomial (or a constant, i.e., a zero-degree polynomial). But because the degree of $p$ is at most two, $q$ must be constant and equal to $0$, and thus $p$ is then the zero polynomial, i.e. the zero vector of the vector space $\mathbb {R} [X]_{\leq 2}$.

Now, since the kernel of $f$ consists only of the zero vector, $f$ is injective.

Proof step: Surjectivity of $f$

In proving this assertion, we use polynomial interpolation in the Lagrangian form.

For this purpose we define three polynomials $p_{1},p_{2},p_{3}\in \mathbb {R} [X]_{\leq 2}$ via

${\begin{aligned}p_{1}(x)&={\dfrac {(x-0)(x-1)}{(-1-0)(-1-1)}}\\p_{2}(x)&={\dfrac {(x-(-1))(x-1)}{(0-(-1))(0-1)}}\\p_{3}(x)&={\dfrac {(x-(-1))(x-0)}{(1-(-1))(1-0)}}\end{aligned}}$

$p_{1}$ has zeros at $x=0$ and $x=1$, and the denominator is the numerator at the position $x=-1$. Hence $p_{1}(-1)=1$, since the numerator and denominator then contain the same number.

Quite analogously, $p_{2}(-1)=p_{2}(1)=0$ and $p_{2}(0)=1$ as well as $p_{3}(-1)=p_{3}(0)=$ and $p_{3}(1)=1$.

Now, if we have any vector $x=(x_{1},x_{2},x_{3})^{T}\in \mathbb {R} ^{3}$, then we define the polynomial $p\in \mathbb {R} [X]_{\leq 2}$ by

$p(x)=x_{1}\cdot p_{1}(x)+x_{2}\cdot p_{2}(x)+x_{3}\cdot p_{3}(x)$

Then, $p(-1)=x_{1}\cdot 1+x_{2}\cdot 0+x_{3}\cdot 0=x_{1}$ ans analogously $p(0)=x_{1}\cdot 0+x_{2}\cdot 1+x_{3}\cdot 0=x_{2}$ as well as $p(1)=x_{1}\cdot 0+x_{2}\cdot 0+x_{3}\cdot 1=x_{3}$.

Thus we have shown that $f$ is surjective.

We also see that we can use this procedure for arbitrary degrees of polynomials and arbitrary points, as long as the number of points is equal to the maximum degree of the polynomials plus 1. We can also replace $\mathbb {R}$ everywhere by $\mathbb {Q}$ or $\mathbb {C}$ without the need to change anything in the proof.

Example (Convergent sequences modulo zero sequences)

Let $K\in \{\mathbb {R} ,\mathbb {C} \}$. We already know the vector spaces $c,c_{0}\subseteq \omega$ of convergent sequences and of zero sequences. We also know that $c_{0}\subseteq c$ is a subspace. Therefore, we can form the factor space $V:=c/c_{0}$.

In the following, we will show that $V\cong K$.

We define a mapping

${\begin{aligned}\Phi \colon K&\to V\\\lambda &\mapsto [(\lambda )_{i\in \mathbb {N} }].\end{aligned}}$

So the image of $\lambda \in K$ under this map is the coset of the sequence which takes the constant value $\lambda$. This is convergent with limit $\lambda$. We have to show that $\Phi$ is linear and bijective.

Proof step: Linearity of $\Phi$

We need to show additivity and homogeneity of $\Phi$ in order to get linearity.

Proof step: Additivity of $\Phi$

Let $\lambda ,\mu \in K$. Then

${\begin{aligned}\Phi (\lambda )+\Phi (\mu )&=[(\lambda )_{i\in \mathbb {N} }]+[(\mu )_{i\in \mathbb {N} }]\\&=[(\lambda )_{i\in \mathbb {N} }+(\mu )_{i\in \mathbb {N} }]\\&=[(\lambda +\mu )_{i\in \mathbb {N} }]\\&=\Phi (\lambda +\mu ).\end{aligned}}$

Proof step: Homogeneity of $\Phi$

Let $\lambda ,\kappa \in K$. We consider $\lambda$ as a vector of the $K$-vector space $K$ and $\kappa$ as a scalar. Then

${\begin{aligned}\Phi (\kappa \cdot \lambda )&=[(\kappa \cdot \lambda )_{i\in \mathbb {N} }]\\&=[\kappa \cdot (\lambda )_{i\in \mathbb {N} }]\\&=\kappa \cdot [(\lambda )_{i\in \mathbb {N} }]\\&=\kappa \cdot \Phi (\lambda ).\end{aligned}}$

Proof step: Injectivity of $\Phi$

We need to show that $\ker(\Phi )=0$ . So let $\lambda \in \ker(\Phi )$. That means, $0=\Phi (\lambda )=[(\lambda )_{i\in \mathbb {N} }]$. Thus, we have $(\lambda )_{i\in \mathbb {N} }\in c_{0}$.

Therefore, $\lambda =\lim _{i\to \infty }\lambda =0$ which establishes the assertion.

Proof step: Surjectivity of $\Phi$

Let $x=[(x_{i})_{i\in \mathbb {N} }]\in V$ and let $\lambda =\lim _{i\to \infty }x_{i}$. We set $y=[(\lambda )_{i\in \mathbb {N} }]=\Phi (\lambda )$. Then $\lim _{i\to \infty }|x_{i}-y_{i}|=\lim _{i\to \infty }|x_{i}-\lambda |=0$.

Therefore $(x_{i})_{i\in \mathbb {N} }-(\lambda )_{i\in \mathbb {N} }\in c_{0}$, which implies $x=y=\Phi (\lambda )$. This establishes surjectivity.

Example (The isomorphism theorem)

One of the most important examples is the isomorphism between the image space of a linear map $L:V\to W$ and the quotient space $V/\ker L$.

All this is described here.

Exercises

Exercise (complex $\mathbb {R}$-vector spaces)

Let $V$ be a finite-dimensional $\mathbb {C}$-vector space. Show that $V\cong \mathbb {R} ^{2\operatorname {dim} _{\mathbb {C} }(V)}$ (interpreted as $\mathbb {R}$-vector spaces).

Solution (complex $\mathbb {R}$-vector spaces)

Set $n:=\operatorname {dim} _{\mathbb {C} }(V)$. We choose a $\mathbb {C}$ basis ${\mathcal {B}}=\{b_{1},\dots ,b_{n}\}$ of $V$. Define $c_{j}:=i\cdot b_{j}$ for all $1\leq j\leq n$.

We have to show that $\{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}$ is an $\mathbb {R}$-basis of $V$. Then, $\operatorname {dim} _{\mathbb {R} }(V)=2n=\operatorname {dim} _{\mathbb {R} }(\mathbb {R} ^{2n})$. According to a theorem above, we have $V\cong \mathbb {R} ^{2n}$ as $\mathbb {R}$-vector spaces.

We now show $\mathbb {R}$-linear independence.

Proof step: $\{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}$ is $\mathbb {R}$-linearly independent

Let $\beta _{1},\dots ,\beta _{n},\gamma _{1},\dots ,\gamma _{n}\in \mathbb {R}$ and assume that $\sum _{j=1}^{n}\beta _{j}\cdot b_{j}+\sum _{j=1}^{n}\gamma _{j}\cdot c_{j}=0$. We substitute the definition for $c_{j}$, conclude the sums and obtain $\sum _{j=1}^{n}(\beta _{j}+i\cdot \gamma _{j})\cdot b_{j}=0$. By $\mathbb {C}$-linear independence of $b_{j}$ we obtain $\beta _{j}+i\cdot \gamma _{j}=0$ for all $j\in \{1,\dots ,n\}$. Thus, $\beta _{j}=\gamma _{j}=0$ for all $j\in \{1,\dots ,n\}$. This establishes the $\mathbb {R}$-linear independence.

Now only one step is missing:

Proof step: $\{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}$ is a generator with respect to $\mathbb {R}$

Let $v\in V$ be arbitrary.

Since ${\mathcal {B}}$ is a $\mathbb {C}$-basis of $V$ , we can find some $\lambda _{1},\dots ,\lambda _{n}\in \mathbb {C}$, such that $v=\sum _{j=1}^{n}\lambda _{j}\cdot b_{j}$ . We write $\lambda _{j}=\beta _{j}+\gamma _{j}i$ with $\beta _{j},\gamma _{j}\in \mathbb {R}$ for all $j$. Then we obtain

${\begin{aligned}v&=\sum _{j=1}^{n}\lambda _{j}\cdot b_{j}\\&=\sum _{j=1}^{n}(\beta _{j}+\gamma _{j}i)\cdot b_{j}\\&=\sum _{j=1}^{n}(\beta _{j}\cdot b_{j}+\gamma _{j}\cdot (i\cdot b_{j}))\\&=\sum _{j=1}^{n}(\beta _{j}\cdot b_{j}+\gamma _{j}\cdot c_{j})\\&=\sum _{j=1}^{n}\beta _{j}\cdot b_{j}+\sum _{j=1}^{n}\gamma _{j}\cdot c_{j}.\end{aligned}}$

So $v$ is inside the $\mathbb {R}$-span of $\{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}$. This establishes the assertion.

Exercise (Isomorphic coordinate spaces)

Let $K$ be a field and consider $n,m\in \mathbb {N} _{0}$. Prove that $K^{n}\cong K^{m}$ holds if and only if $m=n$.

Solution (Isomorphic coordinate spaces)

We know that $\operatorname {dim} (K^{k})=k$ for all $k\in \mathbb {N} _{0}$. We use the theorem above, which states that finite-dimensional vector spaces are isomorphic exactly if their dimensions coincide. So $K^{n}\cong K^{m}$ holds if and only if $n=\operatorname {dim} (K^{n})=\operatorname {dim} (K^{m})=m$ .

Exercise (Isomorphism criteria for endomorphisms)

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space and $f:V\to V$ a $K$-linear map. Prove that the following three statements are equivalent:

(i) $f$ is an isomorphism.

(ii) $f$ is injective.

(iii) $f$ is surjective.

(Note: For this task, it may be helpful to know the terms kernel and image of a linear map. Using the dimension theorem, this exercise becomes much easier. However, we give a solution here, which works without the dimension theorem).

Solution (Isomorphism criteria for endomorphisms)

(i)$\implies$(ii) and (iii): According to the definition of an isomorphism, $f$ is bijective, i.e. injective and surjective. Therefore (ii) and (iii) hold.

(ii)$\implies$(i): Let $f$ be an injective mapping. We need to show that $f$ is also surjective. The image $\mathrm {im} (f):=\{f(v)~|~v\in V\}$ of $f$ is a subspace of $V$. This can be verified by calculation. We now define a mapping $f'$ that does the same thing as $f$, except that it will be surjective by definition. This mapping is defined as follows:

${\begin{aligned}f':V&\to \mathrm {im} (f)\\v&\mapsto f(v)\end{aligned}}$

The surjectivity comes from the fact that every element $w\in \mathrm {im} (f)$ can be written as $w=f(v')$, for a suitable $v'\in V$. Moreover, the mapping $f'$ is injective and linear. This is because $f$ already has these two properties. So $V$ and $\mathrm {im} (f)$ are isomorphic. Therefore, $V$ and $\mathrm {im} (f)$ have the same finite dimension. Since $\mathrm {im} (f)$ is a subspace of $V$, $\mathrm {im} (f)=V$ holds. This can be seen by choosing a basis in $\mathrm {im} (f)$, for instance the basis given by the vectors $v_{1},\dots ,v_{n}\in \mathrm {im} (f)$. These $v_{1},\dots ,v_{n}$ are also linearly independent in $V$, since $\mathrm {im} (f)\subseteq V$. And since $V$ and $\mathrm {im} (f)$ have the same dimension, the $v_{1},\dots ,v_{n}$ are also a basis in $V$. So the two vector spaces $V$ and $\mathrm {im} (f)$ must now be the same, because all elements from them are $K$-linear combinations formed with the $v_{1},\dots ,v_{n}$. Thus we have shown that $f$ is surjective.

(iii)$\implies$(i): Now suppose $f$ is surjective. We need to show that $f$ is also injective. Let $\mathrm {ker} (f):=\{v\in V~|~f(v)=0\}$ be the kernel of the mapping $f$. You may convince yourself by calculation, that this kernel is a subspace of $V$. Let $v_{1},\dots ,v_{k}$ be a basis of $\mathrm {ker} (f)$. We can complete this (small) basis to a (large) basis of $V$, by including the additional vectors $v_{k+1},\dots ,v_{n}$. We will now show that $f(v_{k+1}),\dots ,f(v_{n})$ are linearly independent. So let coefficients $\lambda _{k+1},\dots ,\lambda _{n}\in K$ be given such that

${\begin{aligned}\lambda _{k+1}f(v_{k+1})+\dots +\lambda _{n}f(v_{n})=0.\end{aligned}}$

By linearity of $f$ we conclude: $f(\lambda _{k+1}v_{k+1}+\dots \lambda _{n}v_{n})=0$. This means that the linear combination

${\begin{aligned}\lambda _{k+1}v_{k+1}+\dots +\lambda _{n}v_{n}\end{aligned}}$

is in the kernel of $f$. But we already know a basis of $\mathrm {ker} (f)$. Therefore there are coefficients $\lambda _{1},\dots ,\lambda _{k}\in K$, such that

${\begin{aligned}\lambda _{k+1}v_{k+1}+\dots +\lambda _{n}v_{n}=\lambda _{1}v_{1}+\dots +\lambda _{k}v_{k}.\end{aligned}}$

Because of the linear independence of $v_{1},\dots ,v_{n}$ it now follows that $\lambda _{1},\dots ,\lambda _{n}=0$. Therefore, the $f(v_{k+1}),\dots ,f(v_{n})$ are linearly independent. Next, we will show that these vectors also form a basis of $V$. To do this, we show that each vector in $V$ can be written as a linear combination of the $f(v_{k+1}),\dots ,f(v_{n})$. Let $w\in V$. Because of the surjectivity of $f$, there is a $v\in V$, with $w=f(v)$. Since the $v_{1},\dots ,v_{n}$ form a basis of $V$, there are coefficients $\lambda _{1},\dots ,\lambda _{n}\in K$ such that

${\begin{aligned}v=\lambda _{1}v_{1}+\dots +\lambda _{n}v_{n}\end{aligned}}$

If we now apply $f$ to this equation, we get:

${\begin{aligned}w=f(v)=\lambda _{1}\underbrace {f(v_{1})} _{=0}+\dots +\lambda _{k}\underbrace {f(v_{k})} _{=0}+\lambda _{k+1}f(v_{k+1})+\dots +\lambda _{n}f(v_{n}).\end{aligned}}$

Here we used the linearity of $f$. Since the first $k$ elements of our basis are in the kernel, their images are $0$. So we get the desired representation of $w$:

${\begin{aligned}w=f(v)=\lambda _{k+1}f(v_{k+1})+\dots +\lambda _{n}f(v_{n}).\end{aligned}}$

Thus we have shown that $f(v_{k+1}),\dots ,f(v_{n})$ forms a linearly independent generator of $V$. So these vectors form a basis of $V$. Now if $k$ were not $0$, two finite bases in $V$ would not contain equally many elements. This cannot be the case. Therefore, $k=0$, so $\mathrm {ker} (f)$ is the trivial vector space and $f$ is indeed injective.

Exercise (Function spaces)

Let $X$ be a finite set with $n\in \mathbb {N}$ elements and let $K$ be a field. We have seen that the set of functions from $X$ to $K$ forms a $K$-vector space, denoted by $\operatorname {Fun} (X,K)$. Show that $\operatorname {Fun} (X,K)\cong K^{n}$.

Solution (Function spaces)

We already know according to a theorem above that two finite dimensional vector spaces are isomorphic exactly if they have the same dimension. So we just need to show that $\operatorname {dim} (\operatorname {fun} (X,K))=n=\operatorname {dim} (K^{n})$ holds.

To show this, we first need a basis of $\operatorname {Fun} (X,K)$. For this, let $x_{1},\dots ,x_{n}$ be the elements of the set $X$. We define $f_{1},\dots ,f_{n}\in \operatorname {Fun} (X,K)$ by

$f_{j}(x_{i}):=\delta _{i,j}={\begin{cases}1,{\text{ for }}i=j\\0,{\text{ for }}i\neq j.\end{cases}}$

We now show that the functions $f_{1},\dots ,f_{n}$ indeed form a basis of $\operatorname {Fun} (X,K)$.

Proof step: $f_{1},\dots ,f_{n}$ are linearly independent

Let $\alpha _{1},\dots ,\alpha _{n}\in K$ with $\sum _{k=1}^{n}\alpha _{k}\cdot f_{k}=0$ being the zero function. If we apply this function to any $x_{j}$ with $j\in \{1,\dots ,n\}$, then we obtain: $\sum _{k=1}^{n}\alpha _{k}\cdot f_{k}(x_{j})=0$. By definition of $f_{1},\dots ,f_{n}$ it follows that

$0=\sum _{k=1}^{n}\alpha _{k}f_{k}(x_{j})=\alpha _{j}\cdot f_{j}(x_{j})=\alpha _{j}\cdot 1=\alpha _{j}$.

Since $j$ was arbitrary and $\sum _{k=1}^{n}\alpha _{k}\cdot f_{k}(x_{j})=0$ must hold for all $x_{j}\in X$, it follows that $\alpha _{1}=\dots =\alpha _{n}=0$. So we have shown that $f_{1},\dots ,f_{n}$ are linearly independent.

Proof step: $f_{1},\dots ,f_{n}$ generate $\operatorname {Fun} (X,K)$

Let $g\in \operatorname {Fun} (X,K)$ be arbitrary. We now want to write $g$ as a linear combination of $f_{1},\dots ,f_{n}$. For this we show $g=\sum _{j=1}^{n}g(x_{j})\cdot f_{j}$, i.e., $g$ is a linear combination of $f_{1},\dots ,f_{n}$ with coefficients $g(x_{1}),\dots ,g(x_{n})\in K$. We now verify that $g(x_{i})=\sum _{j=1}^{n}g(x_{j})\cdot f_{j}(x_{i})$ for all $i\in \{1,\dots ,n\}$. Let $i\in \{1,\dots ,n\}$ be arbitrary. By definition of $f_{1},\dots ,f_{n}$ we obtain:

$\sum _{j=1}^{n}g(x_{j})\cdot f_{j}(x_{i})=g(x_{i})\cdot f_{i}(x_{i})=g(x_{i})\cdot 1=g(x_{i})$.

Since equality holds for all $i$, the functions agree at every point and are therefore identical. So we have shown that $f_{1},\dots ,f_{n}$ generate $\operatorname {Fun} (X,K)$.

Thus we have proved that $f_{1},\dots ,f_{n}$ is a basis of $\operatorname {Fun} (X,K)$. Since we have $n$ basis elements of $\operatorname {Fun} (X,K)$, it follows that $\operatorname {dim} (\operatorname {Fun} (X,K))=n=\operatorname {dim} (K^{n})$.

Hint

If $X$ is an infinite set, then $\operatorname {Fun} (X,K)$ is infinite-dimensional. In the special case $X=\mathbb {N}$, $\operatorname {Fun} (X,K)$ is isomorphic to the sequence space $\omega$.