# Serlo: EN: Endomorphism and Automorphism

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An endomorphism is a linear deformation of a vector space $V$. Formally, an endomorphism is a linear mapping $f$ that sends $V$ to itself, i.e., $f\colon V\to V$. A bijective endomorphism is called an automorphism. Intuitively, an automorphism is a linear deformation that can be undone.

## Derivation

We already know linear maps. These are mappings between two vector spaces which are compatible with the (linear) vector space structure. We now examine a few examples of linear maps that we have already learned about in previous articles.

### Examples in the $\mathbb {R} ^{2}$
#### Stretching in $x$-direction

First, we consider the stretching of a vector in the plane by a factor of $2$ in the $x$-direction. Our mapping is thus
$f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)\mapsto (2x,y).$
One can easily verify that $f$ is a linear map. We can illustrate $f$ as follows: We place a checkerboard pattern in the plane and apply $f$ to this checkerboard pattern.
The result is that the boxes are stretched by a factor of $2$ in the $x$-direction.

#### Rotation around the origin

We now consider a rotation $D_{\alpha }$ by the angle $\alpha$ counter-clockwise, with the origin as center of rotation. This is a mapping $D_{\alpha }:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ which assigns to each vector $v\in \mathbb {R} ^{2}$ the vector $D_{\alpha }(v)\in \mathbb {R} ^{2}$ rotated by the angle $\alpha$:
In the introductory article on linear maps we have seen that rotations about the origin are linear. We can visualize $D_{\alpha }$ as in the first example by applying the mapping to the checkerboard pattern. The individual squares then sustain their shape, but they are rotated.

#### Projection on a line

At last we consider the map
$P\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)\mapsto (x,x).$
The map $P$ "presses" vectors onto the straight line $G=\{a\cdot (1,1)^{T}|a\in \mathbb {R} \}$. You can easily check that $P$ is a linear map. We also apply this linear map to the checkerboard pattern to visualize it.
The entire grid is "flattened" onto the straight line $G$.

### Linear deformations of an arbitrary vector space

In all the above examples, we were able to visualize the linear maps as distortions of the checkerboard pattern in $\mathbb {R} ^{2}$. This was possible because all of the above functions map from $\mathbb {R} ^{2}$ into itself. We can illustrate any linear maps $\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ as a deformation of a checkerboard. The deformation of the checkerboard shows us how the map acts on the standard basis vectors $(1,0)^{T}$ and $(0,1)^{T}$ of $\mathbb {R} ^{2}$ and integer multiples of them.
Any linear map $\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ is a linear deformation of the space $\mathbb {R} ^{2}$. Let us generalize this idea to general vector spaces $V$. We can think of linear maps from $V$ to $V$ as linear deformations or transformations of the vector space $V$. In contrast, a linear map $V$ is a transport of the vector space $V$ to $W$. We give a separate name to a linear map which deforms the vector space, i.e., which maps from $V$ to $V$: Such a linear map will be called an endomorphism. So endomorphisms are just those linear maps for which the domain of definition and the target space coincide.

### Reversible deformations

In the examples in $\mathbb {R} ^{2}$ we have seen that some deformations preserve the space and others "flatten" it in a certain sense. The mappings that preserve space can be undone. When the space is flattened, the ma cannot be undone because information is lost. For example, in the above linear map "projection onto a straight line", information is lost about what the $y$-component of the original vector was. It is not possible to recover the vector after applying the transformation. So there are deformations of space that can be undone, and some that cannot. One can undo a deformation exactly if the associated mapping is bijective, i.e., invertible. This gives us the definition of a reversible deformation of space, i.e., an invertible endomorphism. Such a mapping is called an automorphism.

## Definition

Definition (Endomorphism and automorphism)
Let $V$ be a $K$-vector space. A linear map $f\colon V\to V$ that maps $V$ to itself is called an endomorphism. We denote the set of all endomorphisms of $V$ by $\operatorname {End} _{K}(V)$.
We call a bijective endomorphism automorphism. The set of all automorphisms of $V$ is denoted by $\operatorname {Aut} _{K}(V)$.
Hint
Every automorphism is a bijective linear map and therefore also an isomorphism. But not every isomorphism is an automorphism. This is because isomorphisms can also map two different vector spaces to each other.

## Examples
### Examples in $\mathbb {R} ^{2}$
#### Reflection

We consider the linear map $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},(x,y)^{T}\mapsto (-x,y)^{T}$. Since it maps the vector space $\mathbb {R} ^{2}$ into itself, $f$ is an endomorphism. The mapping $f$ preserves $y$ and sends $x$ to $-x$. Thus, we can think of $f$ as a reflection along the $y$-axis. We can undo a reflection by reflecting a second time. This means $f$ is its own inverse mapping. Formally, this is denoted $f\circ f=\operatorname {id} _{\mathbb {R} ^{2}}$ or $f^{-1}=f$. Such a mapping is also called "self-inverse." Because $f$ has an inverse, that is, it is invertible, it follows that $f$ is bijective. Thus, $f$ is also an automorphism.

#### Rotation by 90°

Next we consider the endomorphism $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},(x,y)^{T}\mapsto (-y,x)^{T}$. This is a counter-clockwise rotation by $90$ degrees. To convince yourself that $f$ really is such a rotation, you may calculate how $f$ acts on the standard basis vectors $(1,0)^{T}$ and $(0,1)^{T}$. If it is such a rotation on these two vectors, then by linearity, $f$ must be such a rotation everywhere. A short calculation gives $f((1,0)^{T})=(0,1)^{T}$, as well as $f((0,1)^{T})=(-1,0)^{T}$, which is exactly the desired rotation. Again, we can easily specify an inverse by "turning back" or rotating clockwise by $90$ degrees. This rotation is given by $g\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},(x,y)^{T}\mapsto (y,-x)^{T}$. We briefly check that $g$ is indeed the inverse of $f$:
${\begin{aligned}(f\circ g){\begin{pmatrix}x\\y\end{pmatrix}}&=f{\begin{pmatrix}y\\-x\end{pmatrix}}={\begin{pmatrix}-(-x)\\y\end{pmatrix}}={\begin{pmatrix}x\\y\end{pmatrix}}{\text{, as well as }}\\[0.5em](g\circ f){\begin{pmatrix}x\\y\end{pmatrix}}&=g{\begin{pmatrix}-y\\x\end{pmatrix}}={\begin{pmatrix}x\\-(-y)\end{pmatrix}}={\begin{pmatrix}x\\y\end{pmatrix}}\end{aligned}}$
for $(x,y)^{T}\in \mathbb {R} ^{2}$. So $f\circ g=\operatorname {id} _{\mathbb {R} ^{2}}=g\circ f$ and $f$ is also an automorphism in this example.

#### Shears

Let $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},(x,y)^{T}\mapsto (x+2y,y)^{T}$. The following animation shows how this mapping deforms the plane.
The transformation looks reversible, that is, it looks like an automorphism. We can check this by showing that $f$ is both injective and surjective.
To show injectivity, let us look at the kernel of $f$, i.e., the set $\{(x,y)^{T}\in \mathbb {R} ^{2}\mid f((x,y)^{T})=(0,0)^{T}\}$. For a vector $(x,y)^{T}$ in the kernel, then $(x+2y,y)^{T}=(0,0)^{T}$ holds. From this we directly get $y=0$ and therefore also $x=x+2y=0$. Thus, the kernel consists only of the zero vector and hence $f$ is injective.
To show surjectivity, we take any $(x',y')^{T}\in \mathbb {R} ^{2}$ and find a suitable preimage. That means, we are looking for some $(x,y)^{T}\in \mathbb {R} ^{2}$ with $(x+2y,y)^{T}=f((x,y)^{T})=(x',y')^{T}$. It is directly clear that $y=y'$ must hold. Furthermore, $x+2y=x+2y'=x'$ must be true. This can be transformed to $x=(x'-2y')^{T}$. So $(x'-2y',y')^{T}$ is a preimage of $(x',y')^{T}$. Since $(x',y')^{T}$ was arbitrary, $f$ is surjective.
Linear maps of the form $f((x,y)^{T})=(x+\lambda y,y)^{T}$ with $\lambda \in K$ are called shears. You can show as an exercise that a shear is always an automorphism, no matter what $\lambda$ is.

#### Flattening to the $x$-axis

Let us now consider the mapping $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},(x,y)^{T}\mapsto (x+y,0)^{T}$. This is an endomorphism from $\mathbb {R} ^{2}$ to $\mathbb {R} ^{2}$ that maps every point on the plane to a point on the $x$-axis. So we can think of $f$ as "flattening" the 2-dimensional plane onto the 1-dimensional $x$-axis."
Since $f$ maps the points in $\mathbb {R} ^{2}$ exclusively onto the $x$ axis, $f$ is not a surjective mapping. Nor is it injective, because for every $z\in \mathbb {R}$ we can find different $x,y\in \mathbb {R}$ such that $x+y=z$ holds, e.g., $x=z,y=0$ and vice versa. So $f$ is not an automorphism.

### Example in $\mathbb {R} ^{3}$

Let us now consider an example in $\mathbb {R} ^{3}$. For this we look at the linear map $f\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},(x,y,z)^{T}\mapsto (x+y,x+y,z)^{T}$. Because $f$ maps the vector space $\mathbb {R} ^{3}$ back to $\mathbb {R} ^{3}$, the mapping is an endomorphism.
We now want to check whether $f$ is also an automorphism. To do this, we need to check surjectivity and injectivity. For injectivity, we consider the kernel of $f$, that is, the set $\{(x,y,z)^{T}\in \mathbb {R} ^{3}|f((x,y,z)^{T})=(0,0,0)^{T}\}$. Thus, for vectors $(x,y,z)^{T}$ from the kernel of $f$, $(x+y,x+y,z)^{T}=(0,0,0)^{T}$ holds. From this we can directly conclude that $z=0$ and $x+y=0$, so $y=-x$, must hold. We thus see that the kernel of $f$ contains not only the zero vector, but also the set of all vectors $\{(x,-x,0)^{T}|x\in \mathbb {R} \}$. Hence, $f$ is not injective and therefore cannot be bijective. In particular, $f$ is not an automorphism.
Visually, $f$ compresses vectors onto the plane $\{(x,x,z)^{T}\mid x,z\in \mathbb {R} \}$. Thus, information is lost. Given a vector $(x,x,z)^{T}\in \mathbb {R} ^{3}$, it is no longer possible to say in an unambiguous way from which vector $(a,b,z)^{T}\in \mathbb {R} ^{3}$ it arose under the mapping $f$, since there are very many ways to represent $x\in \mathbb {R}$ as the sum of two numbers $a,b\in \mathbb {R}$. For example, $5=2+3=1+4=-1000+1005$.

### Example in sequence space

There are also endomorphisms on vector spaces other than $\mathbb {R} ^{2}$ and $\mathbb {R} ^{3}$. For instance, we may take an arbitrary field $K$ and consider, as a vector space over it, the sequence space
$\omega :=\{(x_{i})_{i}\mid x_{i}\in K{\text{ for all }}i\in \mathbb {N} \}.$
Now, we take the mapping
${\begin{aligned}f\colon \omega &\to \omega \\[0.3em](x_{i})_{i}&\mapsto f\left((x_{i})_{i}\right)=(x_{k(i)})_{i},\end{aligned}}$
where
$k(i)={\begin{cases}i-1&{\text{if }}i{\text{ is even}}\\i+1&{\text{if }}i{\text{ is odd}}.\end{cases}}$
If we write out the first sequence members, the situation looks like this:
$f(x_{1},x_{2},x_{3},x_{4},\ldots )=(x_{2},x_{1},x_{4},x_{3},\ldots ).$
Thus, the mapping $f$ interchanges even and odd sequence members. We briefly justify why $f$ is linear. Addition and scalar multiplication in the sequence space is understood component-wise, i.e., for $(x_{i})_{i}$ and $(y_{i})_{i}\in \omega$ and $\lambda \in K$ we have
$(x_{i})_{i}+(y_{i})_{i}=(x_{i}+y_{i})_{i}{\text{ and }}\lambda \cdot (x_{i})_{i}=(\lambda \cdot x_{i})_{i}.$
Since $f$ only swaps the order of the components, $f$ is linear. We can also check linearity of $f$ explicitly.
Question: How do you directly show that $f$ is linear?
We prove additivity and homogeneity of $f$. Let $(x_{i})_{i}$ and $(y_{i})_{i}\in \omega$ and $\lambda \in K$. Then
${\begin{aligned}&f((x_{i})_{i}+(y_{i})_{i})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{addition of sequences}}\right.}\\[0.3em]=&f((x_{i}+y_{i})_{i})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]=&(x_{k(i)}+y_{k(i)})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{addition of sequences}}\right.}\\[0.3em]=&(x_{k(i)})_{i}+(y_{k(i)})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]=&f((x_{i})_{i})+f((y_{i})_{i})\end{aligned}}$
and
${\begin{aligned}&f(\lambda \cdot (x_{i})_{i})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{scalar multiplication of sequences}}\right.}\\[0.3em]=&f((\lambda \cdot x_{i})_{i})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]=&(\lambda \cdot x_{k(i)})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{scalar multiplication of sequences}}\right.}\\[0.3em]=&\lambda \cdot (x_{k(i)})_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]=&\lambda \cdot f((x_{i})_{i}).\end{aligned}}$
So $f$ is an endomorphism of $\omega$. Is $f$ also an automorphism? To answer that, we need to verify that $f$ can be undone. The mapping swaps even and odd sequence members. So if we swap the sequence members back, $f$ is undone. This second swap can be done by just applying $f$ again. As with the very first example, $f$ is self-inverse - in formulas this is called $f\circ f=\operatorname {id} _{\omega }$ or $f=f^{-1}$. Since $f$ is invertible, the mapping is bijective. So $f$ is an automorphism.

## Endomorphisms form a ring with unit

In the article Vector space of a linear map we saw that the set of linear maps $\operatorname {Hom} _{K}(V,W)$ between two $K$-vector spaces $V$ and $W$ again forms a vector space. Since $\operatorname {End} _{K}(V)=\operatorname {Hom} _{K}(V,V)$ holds, the set of endomorphisms is also a vector space. That is, we can add endomorphisms of a vector space $V$ and multiply them by scalars. In particular, we can link two endomorphisms $f$ and $g\in \operatorname {End} _{K}(V)$ by addition and obtain an endomorphism $f+g\in \operatorname {End} _{K}(V)$. This space is given by
$f+g:V\to V,\quad (f+g)(v):=f(v)+_{V}g(v)$
where $+_{V}$ denotes addition in the vector space $V$.
Can $f$ and $g$ be connected in another way? Intuitively, $f$ and $g$ are two representations of the vector space $V$. We can now deform the space $V$ with $f$ and then deform the result with $g$. This produces a new deformation of the vector space. That is, we again get an endomorphism of $V$. This resulting mapping, that corresponds to applying $f$ and $g$ one after the other, is called composition and denoted $g\circ f$. Thus, the composition of two endomorphisms is always an endomorphism. In summary, we can "connect" two endomorphisms $f$ and $g$ by forming the addition $f+g$ or the composition $g\circ f$.
Because we have composition as an operation in addition to addition, $\operatorname {End} _{K}(V)$ carries more structure than just vector space structure. We will prove later that the set $\operatorname {End} _{K}(V)$ of endomorphisms on $V$ forms a ring with these two operations. Here, the addition in the ring is the addition of the mappings and the multiplication in the ring is the composition of the mappings.
It is now an interesting question, when the ring $\operatorname {End} _{K}(V)$ has a unit element and is commutative. If this was the case, we would even have a field and could build more vector spaces over it. Now, a unit element exists if there is a neutral element of multiplication. That is, if there is a $k\in \operatorname {End} _{K}(V)$ such that $k\circ f=f$ and $f\circ k=f$ holds for all $f\in \operatorname {End} _{K}(V)$. We already know a mapping that satisfies this property: the identity $\operatorname {id} _{V}$. This is a linear map $V\to V$ and so $\operatorname {id} _{V}\in \operatorname {End} _{K}(V)$ holds. So the ring $\operatorname {End} _{K}(V)$ has a unit element.
Is $\operatorname {End} _{K}(V)$ a commutative ring? To answer this, we need to check that $f\circ g=g\circ f$ holds for all $f,g\in \operatorname {End} _{K}(V)$. To get an intuition whether the statement is true or false, we consider again examples with $\mathbb {R} ^{2}$. Let $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the projection onto the $y$-axis; that is, for $(x,y)\in \mathbb {R} ^{2}$, $f(x,y)=(0,y)$ holds. Furthermore, let $g\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the rotation by $90^{\circ }$ clockwise (or by $270^{\circ }$ counter-clockwise) about the origin; that is, $g(x,y)=(-y,x)$ holds. We want to investigate whether $f\circ g=g\circ f$. What do the maps $f\circ g$ and $g\circ f$ do visually? The map $g\circ f$ first pushes all space onto the $y$-axis and then rotates it by $90^{\circ }$ in a clockwise direction. So our result is on the $x$-axis.
The mapping $f\circ g$ first rotates the space by $90^{\circ }$ clockwise and then pushes everything to the $y$-axis. So the result of the mapping lies on the $y$-axis.
Consequently, $f\circ g$ and $g\circ f$ are different mappings. Therefore, $\operatorname {End} _{\mathbb {R} }(\mathbb {R} ^{2})$ is not a commutative ring. More generally, for any vector space $V$ with $\dim V\geq 2$, $\operatorname {End} _{K}(V)$ is not a commutative ring. We deal with this below in a corresponding exercise.
As announced above, we now prove that $(\operatorname {End} _{K}(V),+,\circ )$ is always a ring:
Theorem (Endomorphism ring)
Let $K$ be a field and $V$ be a $K$-vector space. Then the set of endomorphisms $\operatorname {End} _{K}(V):=\{f\colon V\to V\mid f{\text{ is linear}}\}$ on $V$ together with addition $+$ and composition $\circ$ a ring with unit.
For two endomorphisms $f,g\in \operatorname {End} _{K}(V)$ , the endomorphisms $f+g\colon V\to V$ and $f\circ g\colon V\to V$ are defined by
$(f+g)(v):=f(v)+_{V}g(v)$ and $(f\circ g)(v):=f(g(v))$ for $v\in V$
Here, $+_{V}$ is the addition on $V$.
Proof (Endomorphism ring)
In order that $(\operatorname {End} _{K}(V),+,\circ )$ forma a ring with unit, the following properties must be satisfied:
* The tuple ($\operatorname {End} _{K}(V),+$) is an Abelian group.
** Associative law of addition: $\forall f,g,h\in \operatorname {End} _{K}(V):\ f+(g+h)=(f+g)+h$
** Commutative law of addition: $\forall f,g\in \operatorname {End} _{K}(V):\ f+g=g+f$
** Existence of an additive neutral element: $\exists e\in \operatorname {End} _{K}(V)\;\forall f\in {\operatorname {End} _{K}(V)}:\ e+f=f$
** Existence of additive inverse: $\forall f\in \operatorname {End} _{K}(V)\;\exists f'\in \operatorname {End} _{K}(V):\ f'+f=e$
* The tuple ($\operatorname {End} _{K}(V),\circ$) is a monoid
** Associative law of multiplication: $\forall f,g,h\in \operatorname {End} _{K}(V):\ f\circ (g\circ h)=(f\circ g)\circ h$
** Existence of a multiplicative neutral element: $\exists k\in \operatorname {End} _{K}(V)\;\forall f\in \operatorname {End} _{K}(V):\ k\circ f=f$
* The distributive laws hold
** Distributive law I: $\forall f,g,h\in \operatorname {End} _{K}(V):\ f\circ (g+h)=f\circ g+f\circ h$
** Distributive law II: $\forall f,g,h\in \operatorname {End} _{K}(V):\ (f+g)\circ h=f\circ h+g\circ h$
Before we start with the proof, let's keep the following simple fact in mind:
Let $f,g,h\in \operatorname {End} K(V)$ and $x\in V$. The mappings $f,g$ and $h$ map elements of $V$ to elements of $V$. Accordingly, $f(x),g(x),h(x)$ are elements of $V$. By premise, $(V,+_{V},\cdot _{V})$ is a $K$ vector space. Therefore, we can apply the computational rules that hold in the $K$-vector space $V$ to the elements $f(x),g(x),h(x)$.
Proof step: ($\operatorname {End} _{K}(V),+$) is an Abelian group
Proof step: Associative law of addition
The associative law of addition reds: $\forall f,g,h\in \operatorname {End} _{K}(V):\ f+(g+h)=(f+g)+h$
We prove this equation by establishing for all $f,g,h\in \operatorname {End} _{K}(V)$ the equality $(f+(g+h))(x)=((f+g)+h)(x)$ for each vector $x\in V$.
So let $f,g,h\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]((f+g)+h)(x)&=(f+g)(x)+_{V}h(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=(f(x)+_{V}g(x))+_{V}h(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associative law of addition in }}V\right.}\\[0.3em]&=f(x)+_{V}(g(x)+_{V}h(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=f(x)+_{V}(g+h)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=(f+(g+h))(x)\\[0.3em]\end{aligned}}$
This shows the associative law of addition.
Proof step: Commutative law of addition
The commutative law of addition is: $\forall f,g\in \operatorname {End} _{K}(V):\ f+g=g+h$ We prove this by establishing that for all $f,g\in \operatorname {End} _{K}(V)$ we have $(f+g)(x)=(g+f)(x)$ for each $x\in V$.
So let $f,g\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em](f+g)(x)&=f(x)+_{V}g(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of addition }}V\right.}\\[0.3em]&=g(x)+_{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=(g+f)(x)\\[0.3em]\end{aligned}}$
This shows the commutativity of the addition.
Proof step: Existence of an additive neutral element
We need to show the following statement: $\exists e\in \operatorname {End} _{K}(V)\;\forall f\in {\operatorname {End} _{K}(V)}:\ e+f=f$
We prove this statement by establishing: $\exists e\in \operatorname {End} _{K}(V)\;\forall f\in {\operatorname {End} _{K}(V)}:\ (e+f)(x)=f(x)\quad x\in V$
For this, let us choose $e=0_{\operatorname {Fun(V,V)} }\in \operatorname {End} K(V)\subseteq \operatorname {Fun} (V,V)$, where $0_{\operatorname {Fun(V,V)} }:V\rightarrow V,x\mapsto 0_{V}$ is the zero map from $V$ to $V$. We now show that $0_{\operatorname {Fun(V,V)} }$ is the neutral element of the addition. For this, let $f\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em](0_{\operatorname {Fun(V,V)} }+f)(x)&=0_{\operatorname {Fun(V,V)} }(x)+_{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}0_{\operatorname {Fun(V,V)} }\right.}\\[0.3em]&=0_{V}+_{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Neutral element of addition in }}V\right.}\\[0.3em]&=f(x)\\[0.3em]\end{aligned}}$
The additive neutral element here is therefore the zero mapping $0_{\operatorname {Fun(V,V)} }:V\rightarrow V,x\mapsto 0_{V}$.
Proof step: Existence of additive inverses
We have to show the following statement: $\forall f\in \operatorname {End} K(V)\;\exists f'\in \operatorname {End} _{K}(V):\ f'+f=0_{\operatorname {Fun(V,V)} }$
This is done by establishing: $\forall f\in \operatorname {End} K(V)\;\exists f'\in \operatorname {End} _{K}(V)\;\forall x\in V:\ (f'+f)(x)=0_{V}=0_{\operatorname {Fun(V,V)} }(x)$
Since $V$ is a $K$-vector space, any vector $v\in V$ has an additive inverse, namely $-v$. Then $v+_{V}(-v)=0_{V}$ holds. Therefore, for any endomorphism $f\in \operatorname {End} _{K}(V)$, we can simply choose $f':V\rightarrow V,x\mapsto -f(x)$. Now we still have to show that. For this, let $f\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em](f'+f)(x)&=f'(x)+_{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f'\right.}\\[0.3em]&=-f(x)+_{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additive inverse in }}V\right.}\\[0.3em]&=0_{V}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}0_{\operatorname {Fun(V,V)} }\right.}\\[0.3em]&=0_{\operatorname {Fun(V,V)} }(x)\\[0.3em]\end{aligned}}$
So the additive inverse of a mapping $f:V\rightarrow V,x\mapsto f(x)$ is given by $f':V\rightarrow V,x\mapsto -f(x)$.
We have thus proved that $(\operatorname {End} _{K}(V),+)$ is an Abelian group.
We could have shown this statement differently. In the article Vector space of a linear map we considered the set of linear maps between two $K$-vector spaces $V$ and $W$. We call this set $\operatorname {Hom} (V,W)$. We have seen that $\operatorname {Hom} (V,W)$ forms a $K$-vector space. It is then true that $\operatorname {End} _{K}(V)=\operatorname {Hom} (V,V)$. So $\operatorname {End} _{K}(V)$ is also a vector space and thus an Abelian group.
Proof step: ($\operatorname {End} _{K}(V),\circ$) is a monoid
Proof step: Associative law of multiplication
The associative law of multiplication in $\operatorname {End} _{K}(V)$ is: $\forall f,g,h\in \operatorname {End} _{K}(V):\ f\circ (g\circ h)=(f\circ g)\circ h$
This is true because the composition of mappings is associative.
Proof step: Existence of a multiplicative neutral element
We have to establish the following statement: $\exists k\in \operatorname {End} _{K}(V)\;\forall f\in {\operatorname {End} _{K}(V)}:\ k\circ f=f$
This is proven by establishing the following statement: $\exists k\in \operatorname {End} _{K}(V)\;\forall f\in {\operatorname {End} _{K}(V)}\;\forall x\in V:\ (k\circ f)(x)=f(x)$
We choose $k=\operatorname {id} _{V}\in \operatorname {End} K(V)\subseteq \operatorname {Fun} (V,V)$, where $\operatorname {id} _{V}:V\rightarrow V,\ x\mapsto x$ is the identity on $V$. We further want to show that $k$ is the neutral element of the multiplication. For this, let $f\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}k\right.}\\[0.3em](k\circ f)(x)&=(\operatorname {id} _{V}\circ f)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\circ \right.}\\[0.3em]&=\operatorname {id} _{V}(f(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of}}\operatorname {id} _{V}\right.}\\[0.3em]&=f(x)\\[0.3em]\end{aligned}}$
So the neutral element of the multiplication in given by the identity on $V$, i.e., $\operatorname {id} _{V}:V\to V,\ x\mapsto x$.
Proof step: Distributive laws
Proof step: Distributive law I
The distributive law I reads: $\forall f,g,h\in \operatorname {End} _{K}(V):\ f\circ (g+h)=f\circ g+f\circ h$
We prove this equation by establishing for all $f,g,h\in \operatorname {End} _{K}(V)$ the equality $(f\circ (g+h))(x)=(f\circ g+f\circ h)(x)$ with $x\in V$. For this, let $f,g,h\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of the composition}}\right.}\\[0.3em](f\circ (g+h))(x)&=f((g+h)(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=f(g(x)+_{V}h(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{f is linear}}\right.}\\[0.3em]&=f(g(x))+_{V}f(h(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of the composition}}\right.}\\[0.3em]&=(f\circ g)(x)+_{V}(f\circ h)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=(f\circ g+f\circ h)(x)\end{aligned}}$
This establishes the distributive law I.
Proof step: Distributive law II
The distributive law II reads: $\forall f,g,h\in \operatorname {End} _{K}(V):\ (f+g)\circ h=f\circ h+g\circ h$
We prove this equation by establishing the equation $((f+g)\circ h)(x)=(f\circ h+g\circ h)(x)$ for all $f,g,h\in \operatorname {End} _{K}(V)$ and $x\in V$. So let $f,g,h\in \operatorname {End} K(V)$ and $x\in V$. Then
${\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of the composition}}\right.}\\[0.3em]((f+g)\circ h)(x)&=(f+g)(h(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=f(h(x))+_{V}g(h(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of the composition}}\right.}\\[0.3em]&=(f\circ h)(x)+_{V}(g\circ h)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}+\right.}\\[0.3em]&=(f\circ h+g\circ h)(x)\end{aligned}}$
This establishes the distributive law II.

## Automorphisms and flattening
### The finite-dimensional case

Above we have already examined some examples of endomorphisms and automorphisms. We have seen that endomorphisms which "flatten" a vector space are not bijective and therefore not automorphisms. On the other hand, endomorphisms which do not "flatten" a vector space are indeed automorphisms.
Question: What does "not flattening" mean in a mathematical language?
An endomorphism $f\colon V\to V$ "does not flatten a vector space" if there is no vector $v\neq 0$ mapped to zero by $f$. That is, for all vectors $v\neq 0$, $f(v)\neq 0$ holds. Since $f$ is a linear map, this is exactly the case if $f$ is injective, i.e., if it is a monomorphism.
For endomorphisms of finite-dimensional vector spaces, being "non-flattening" is equivalent to being an automorphism: Let $f\colon V\to V$ be an endomorphism of an $n$-dimensional vector space $V$. If the mapping $f$ is an automorphism, then it is injective. So $f$ does not flatten $V$. Conversely, if we assume that $f$ does not flatten $V$, it follows that $f$ is injective. Thus, no information from $V$ is lost when mapping with $f$. From this, we can conclude that the image $\operatorname {im} (f)=f(V)$ is also $n$-dimensional. So $\operatorname {im} (f)=V$ must hold. Thus, $f$ is also surjective and therefore an automorphism.
We have seen that an injective endomorphism over a finite-dimensional vector space is automatically surjective. Does the converse statement also hold? In other words: If $f\colon V\to V$ is a surjective endomorphism of a $n$-dimensional vector space, does it follow that $f$ is injective? If $f$ is surjective, then $f(V)=V$ and hence $\dim(f(V))=n$ holds. Suppose $f$ is not injective. Then there is a vector $0\neq v\in V$ for which $f(v)=0$. Thus, $f$ "flattens the direction" in which $v$ points. This means, when mapping $V$ by $f$, we lose at least one dimension of $V$. Consequently, we would have $\dim(f(V))<n$. This is a contradiction to $\dim(f(V))=n$. Therefore, $f$ must be injective. So if $f$ is surjective, then $f$ is also injective.
possibly refer to an explanation in the article "Linear maps between finite dimensional vector spaces" when it is written.
We show these statements again formally in the following theorem.
Theorem (Endomorphisms on finite-dimensional vector spaces)
Let $V$ be a finite-dimensional vector space and $f\colon V\to V$ be an endomorphism. Then, the following statements are equivalent
* $f$ is an isomorphism
* $f$ is a monomorphism
* $f$ is an epimorphism
Proof (Endomorphisms on finite-dimensional vector spaces)
We already know that for two finite-dimensional vector spaces $V$ and $W$ with $\dim V=\dim W$ and a linear map $f\colon V\to W$, that the three statements
* $f$ is an isomorphism
* $f$ is a monomorphism
* $f$ is an epimorphism
are equivalent. So for an endomorphism $f\colon V\to V$ from a finite-dimensional vector space $V$, the three statements must also be equivalent.
Link the theorem for general linear maps as soon as it is written

### The infinite-dimensional case

In the infinite-dimensional case, the above argument no longer works. We have exploited in the finite-dimensional case that for an $n$-dimensional vector space $V$ and a subspace $U\subseteq V$ it already follows from $\dim(U)=n$ that $V=U$. Above, we used $U=f(V)$. However, in infinite-dimensional vector spaces this does not hold. Here, a paradoxical effect occurs: One can place an infinite-dimensional subspace $U$ into another infinite-dimensional space $V$ of the same size, without filling all of $V$ (This is related to Hilbert's Hotel paradox).
So for endomorphisms $f\colon V\to V$ of an infinite-dimensional vector space $V$, it does not hold that $f$ is surjective exactly when $f$ is injective. To understand this better, we now examine concrete counterexamples.
Example (An injective endomorphism that is not surjective)
Let $V=\{(x_{i})_{i}\mid x_{i}\in \mathbb {R} {\text{ for all }}i\in \mathbb {N} \}$ be the sequence space over $\mathbb {R}$. We define the endomorphism
$f\colon V\to V,\quad (x_{1},x_{2},x_{3},\ldots )\mapsto (0,x_{1},x_{2},x_{3},\ldots ).$
You can easily verify that $f$ is linear. Why is $f$ injective? For $x=(x_{1},x_{2},x_{3},\ldots )\in V$ with $f(x)=0$ we have $0=x_{1}=x_{2}=x_{3}=\ldots$, so $x=0$. Thus, $\ker(f)=\{0\}$ follows and $f$ is injective.
Why is $f$ not surjective? To see this, we need to find a vector in $V$ that is not "hit" by $f$. For instance, consider $(1,0,0,\ldots )\in V$. No matter which $x\in V$ we choose, it holds for $f(x)$ that the first sequence member is equal to $0$. So $(1,0,0,\ldots )$ is never hit by $f$. Therefore, $f$ is not surjective.
Hint
The procedure in this example is an example often given in the context of the Hilbert Hotel paradox. In this example, one shifts all elements of the set $\mathbb {N}$ by $1$ by the mapping
${\begin{aligned}g\colon \mathbb {N} &\to \mathbb {N} \\n&\mapsto n+1.\end{aligned}}$
This mapping is also injective, but just like $f$ it is not surjective. The difference is that $f$ is a linear map between vector spaces and $g$ is only a map on $\mathbb {N}$. So infinite-dimensional vector spaces show some similar weird effects as infinitely large sets.
Example (A surjective endomorphism that is not injective)
We consider again the sequence space $V=\{(x_{i})_{i}\mid x_{i}\in \mathbb {R} {\text{ for all }}i\in \mathbb {N} \}$ over $\mathbb {R}$. Now we define the endomorphism
$f\colon V\to V,\quad (x_{1},x_{2},x_{3},\ldots )\mapsto (x_{2},x_{3},x_{4},\ldots ).$
So $f(x)_{i}=x_{i+1}$ for $x=(x_{i})_{i}\in V$. Again, one can easily verify that $f$ is linear.
First we verify that $f$ is surjective. For this, let $y=(y_{1},y_{2},y_{3}\ldots )\in V$ be any vector. We want to find a vector $x\in V$ for which $f(x)=y$ holds. This is true for $x=(0,y_{1},y_{2},y_{3},\ldots )\in V$ and thus $f$ is surjective.
Why is $f$ not injective? To see this, we need to find some $x\in V$ with $f(x)=0$ and $x\neq 0$. An example is (again) given by $x=(1,0,0,\ldots )\in V$. Then $x\neq (0,0,0,\ldots )=0$, but $f(x)=(0,0,0,\ldots )=0$. Thus, $f$ is not injective.

### The automorphism group

We know that the endomorphisms form a ring with unit. The automorphisms are exactly all invertible endomorphisms. In other words, the automorphisms of a vector space are exactly the multiplicatively invertible elements, of the endomorphism ring. Recall that the multiplication in the endomorphism ring is just the composition $\circ$ of mappings. In the following theorem we show that $\operatorname {Aut} _{K}(V)$ is indeed a group with respect to this multiplication.
Theorem (Automorphism group)
Let $K$ be a field and $V$ a $K$-Vector space. The set $\operatorname {Aut} _{K}(V):=\{\,f\colon V\to V\,|\,f{\text{ automorphism}}\,\}$ forms a group with respect to composition $\circ$.
Further, we have $\operatorname {Aut} _{K}(V)=\operatorname {End} _{K}(V)^{\times }$.
Proof (Automorphism group)
We need to show that
* $\operatorname {Aut} _{K}(V)$ is closed with respect to $\circ$
* $\circ$ is associative
* There is a neutral element with respect to $\circ$ in $\operatorname {Aut} _{K}(V)$
* Each element $f$ in $\operatorname {Aut} _{K}(V)$ has a multiplicative inverse.
Proof step: Closedness with respect to $\circ$
We prove that for all automorphisms $f$ and $g\in \operatorname {Aut} _{K}(V)$ it also holds that $f\circ g\in \operatorname {Aut} _{K}(V)$.
For this, let $f$ and $g\in \operatorname {Aut} _{K}(V)$ . So $f$ and $g$ are also endomorphisms. Because $\operatorname {End} _{K}(V)$ forms a ring with multiplication $\circ$, $\operatorname {End} _{K}(V)$ is closed with respect to $\circ$. Thus, $f\circ g\in \operatorname {End} _{K}(V)$ holds. So all we have to do is to justify that $f\circ g$ is bijective. Because $f$ and $g$ are bijective, there are respectively inverses $f^{-1}\colon V\to V$ and $g^{-1}\colon V\to V$. We now show that $g^{-1}\circ f^{-1}\colon V\to V$ is the inverse of $f\circ g$. It holds that
$(f\circ g)\circ (g^{-1}\circ f^{-1})=f\circ (g\circ g^{-1})\circ f^{-1}=f\circ \operatorname {id} _{V}\circ f^{-1}=f\circ f^{-1}=\operatorname {id} _{V}$
and
$(g^{-1}\circ f^{-1})\circ (f\circ g)=g^{-1}\circ (f^{-1}\circ f)\circ g=g^{-1}\circ \operatorname {id} _{V}\circ g=g^{-1}\circ g=\operatorname {id} _{V}.$
So $f\circ g$ has an inverse and is therefore bijective. Therefore, $f\circ g$ is an automorphism.
Proof step: Associativity of $\circ$
We need to show that for all automorphisms $f,g$ and $h\in \operatorname {Aut} _{K}(V)$ the following holds
$f\circ (g\circ h)=(f\circ g)\circ h.$
Let $f,g$ and $h\in \operatorname {Aut} _{K}(V)$ , as well as $v\in V$ be any vector. Then
${\begin{aligned}&(f\circ (g\circ h))(v)\\=&f((g\circ h)(v))\\=&f(g(h(v)))\\=&(f\circ g)(h(v))\\=&((f\circ g)\circ h)(v).\end{aligned}}$
Since the mappings $f\circ (g\circ h)$ and $(f\circ g)\circ h$ coincide on all vectors $v\in V$, they are equal, i.e., $f\circ (g\circ h)=(f\circ g)\circ h$.
Proof step: Existence of a neutral element
We need to find an element $e\in \operatorname {Aut} _{K}(V)$ such that for all $f\in \operatorname {Aut} _{K}(V)$ it is true that $f\circ e=f=e\circ f$.
We choose $e:=\operatorname {id} _{V}$. Then $e\colon V\to V$ , so $e$ is linear. Thus, $e\in \operatorname {End} _{K}(V)$. Further, $e=\operatorname {id} _{V}$ is bijective, so $e\in \operatorname {Aut} _{K}(V)$.
Let now $f\in \operatorname {Aut} _{K}(V)$ . Then
$f\circ e=f\circ \operatorname {id} _{V}=f=\operatorname {id} _{V}\circ f=e\circ f.$
So $e$ is the neutral element in $\operatorname {Aut} _{K}(V)$.
Proof step: Existence of a multiplicative inverse
We prove that for every $f\in \operatorname {Aut} _{K}(V)$ there exists an automorphism $g\in \operatorname {Aut} _{K}(V)$ such that $f\circ g=\operatorname {id} _{V}=g\circ f$.
Let $f\in \operatorname {Aut} _{K}(V)$ be an automorphism. Then $f$ is bijective and thus there exists an inverse mapping $f^{-1}\colon V\to V$, for which $f\circ f^{-1}=\operatorname {id} _{V}=f^{-1}\circ f$. We only have to show that $f^{-1}\in \operatorname {Aut} _{K}(V)$. What we know is that inverses of linear maps are again linear. Consequently, $f^{-1}$ , as the inverse of the linear map $f$ , is also linear. Furthermore, $f^{-1}$ is bijective because $f^{-1}$ has an inverse $f$. So $f^{-1}\in \operatorname {Aut} _{K}(V)$ holds, and thus $f^{-1}$ is an inverse of $f$ in $\operatorname {Aut} _{K}(V)$.
The automorphisms form a group, but are no longer a ring. This is because $\operatorname {Aut} _{K}(V)$ no longer has an additive structure: If we have two automorphisms $f$ and $g$ from a vector space $V$, $f+g$ need not be an automorphism again. And indeed, there are counterexamples for this:
Example (Sum of automorphisms which is not an automorphism)
We consider the vector space $V=\mathbb {R} ^{2}$ and define the automorphisms
${\begin{aligned}&f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad &g\colon \mathbb {R} ^{2}&\to \mathbb {R} ^{2}\\[0.3em]&{\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x\\y\end{pmatrix}},&{\begin{pmatrix}x\\y\end{pmatrix}}&\mapsto {\begin{pmatrix}x\\-y\end{pmatrix}}.\end{aligned}}$
It is easy to show that $f$ and $g$ are linear and bijective, so $f,g\in \operatorname {Aut} _{\mathbb {R} }(\mathbb {R} ^{2})$. Now
${\begin{aligned}f+g\colon \mathbb {R} ^{2}&\to \mathbb {R} ^{2}\\[0.3em]{\begin{pmatrix}x\\y\end{pmatrix}}&\mapsto {\begin{pmatrix}2x\\0\end{pmatrix}}.\end{aligned}}$
Since this mapping does not hit the vector $(0,1)^{T}$, it is not surjective. So $f+g$ is not bijective and therefore not an automorphism.
Hint
$\operatorname {Aut} _{K}(V)$ is never closed under addition, unless $V=\{0\}$.
For vector spaces $V$ with $\dim V\geq 2$ the automorphism group is not commutative. As with the endomorphism ring, the composition of the mappings is not commutative. We demonstrate this non--commutativity in an exercise below.

## Exercises

Exercise (Automorphism)
Show that $f\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n},\,(x_{1},x_{2},\ldots ,x_{n})^{T}\mapsto (-2x_{1},-2x_{2},\ldots ,-2x_{n})^{T}$ is an automorphism.
Solution (Automorphism)
Linearity can easily be verified. Since the domain and target space are the same, $f$ is therefore an endomorphism.
We now want to show that $f$ is bijective. To do this, we must show that $f$ is injective and surjective.
We start with injectivity. Let $x=(x_{1},\ldots ,x_{n})^{T}$ and $y=(y_{1},\ldots ,y_{n})^{T}\in \mathbb {R} ^{n}$ with $f(x)=f(y)$. Then, for $i=1,\ldots ,n$ we have $-2x_{i}=-2y_{i}$, which implies $x_{i}=y_{i}$ and thus $x=y$. This establishes injectivity.
Now we show surjectivity. For this, let $x=(x_{1},\ldots ,x_{n})^{T}\in \mathbb {R} ^{n}$. We define $y:=(-x_{1}/2,\ldots ,-x_{n}/2)^{T}\in \mathbb {R} ^{n}$. Then $f(y)=x$. So $f$ is indeed surjective.
So we have shown that $f$ is an automorphism.
Exercise (Transformation in the space of Fibonacci sequences)
Let $K$ be a field and $V$ be the vector space of Fibonacci sequences
$V=\{(x_{i})_{i\in \mathbb {N} }\in \omega \mid x_{n}+x_{n+1}=x_{n+2}{\text{ for all }}n\in \mathbb {N} \},$
where $\omega$ is the space of all sequences in $K$. Show:
* $V$ is isomorphic to $K^{2}$.
* There is an endomorphism $f\colon V\to V$ that swaps the first two entries of each sequence, that is, $f((x_{i})_{i\in \mathbb {N} })_{1}=x_{2}$ and $f((x_{i})_{i\in \mathbb {N} })_{2}=x_{1}$ holds for all $(x_{i})_{i\in \mathbb {N} }\in V$.
* $f$ is an automorphism.
Solution (Transformation in the space of Fibonacci sequences)
Proof step: $V\cong K^{2}$
We show that
$\Phi \colon V\to K^{2},\quad (x_{i})_{i\in \mathbb {N} }\mapsto {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}$
is an isomorphism. The linearity can be easily verified.
For injectivity we show $\ker \Phi =\{0\}$. So let $(x_{i})_{i\in \mathbb {N} }\in V$ with $\Phi ((x_{i})_{i\in \mathbb {N} })=(0,0)^{T}$, i.e., $x_{1}=x_{2}=0$. We show $x_{n}=0$ for all $n\in \mathbb {N}$ by induction. By assumption, the statement holds for $n=1,2$. Now let $n>2$ be fixed. To establish the induction step, we must show that $x_{n}=0$. As an induction assumption, we can use that $x_{i}=0$ for all $i\in \{1,2,\ldots ,n-1\}$. By definition of the sequence $(x_{i})_{i\in \mathbb {N} }\in V$ it follows that $x_{n}=x_{n-1}+x_{n-2}=0+0=0$, which establishes the induction step and completes the induction.
For surjectivity, we use that any sequence in $V$ can be defined by specifying the first two members of the sequence: Let $(a,b)^{T}\in K^{2}$. We define $(x_{i})_{i\in \mathbb {N} }$ inductively as in the proof of injectivity for $x_{1}=a$, $x_{2}=b$ and $x_{i}:=x_{i-1}+x_{i-2}$. Then $(x_{i})_{i\in \mathbb {N} }\in V$ and $\Phi ((x_{i})_{i\in \mathbb {N} })=(a,b)^{T}$.
Proof step: There is an endomorphism $f\colon V\to V$ that swaps the first two entries of each sequence.
We use the isomorphism $\Phi$ from the first part of the exercise. Obviously
$g\colon K^{2}\to K^{2},\quad {\begin{pmatrix}a\\b\end{pmatrix}}\mapsto {\begin{pmatrix}b\\a\end{pmatrix}}$
is linear and maps from $K^{2}$ to $K^{2}$, so it is an endomorphism. Thus, $f:=\Phi ^{-1}\circ g\circ \Phi$ is also linear as a concatenation of linear maps. Since $f$ maps from $V$ to $V$, $f$ is an endomorphism, and by construction
$\Phi (f((x_{i})_{i\in \mathbb {N} }))=g(\Phi ((x_{i})_{i\in \mathbb {N} }))=g{\Bigl{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}{\Bigr )}={\begin{pmatrix}x_{2}\\x_{1}\end{pmatrix}}$
for all $(x_{i})_{i\in \mathbb {N} }\in V$. So $f$ swaps the first two entries of each sequence.
Proof step: $f$ is an automorphism.
We have to show that $f$ is an isomorphism. Since $\Phi$ is an isomorphism, $f=\Phi ^{-1}\circ g\circ \Phi$ is an isomorphism if and only $g$ is also an isomorphism. The endomorphism $g$ simply swaps the two components of a vector in $K^{2}$. So it maps the (ordered) basis $(e_{1},e_{2})$ from $K^{2}$ to the basis $(e_{2},e_{1})$. Since a linear map is bijective if and only if it maps bases to bases, $g$ is an isomorphism.
Exercise (Shears are automorphisms)
Let $\lambda \in \mathbb {R}$ be a scalar. We consider the mapping $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (x+\lambda y,y)^{T}$. Show that $f$ is an automorphism.
Solution (Shears are automorphisms)
The linearity of $f$ can easily be verified. Since $f$ maps $\mathbb {R} ^{2}$ to itself, $f$ is an endomorphism and all that remains is to show bijectivity.
We prove injectivity by showing $\ker(f)=\{(0,0)^{T}\}$. Let $u=(u_{1},u_{2})^{T}\in \ker(f)$, that is, $f(u)=(0,0)^{T}$ holds. We want to show $u=(0,0)^{T}$. Since $f(u)=(u_{1}+\lambda u_{2},u_{2})^{T}=(0,0)^{T}$ holds, we get $u_{2}=0$ from the second vector component. It now follows that $0=u_{1}+\lambda \cdot 0=u_{1}$ and that $u=(0,0)$ holds. Thus, $f$ is injective.
Second, we need to show that $f$ is surjective. For this, let $v=(v_{1},v_{2})^{T}\in \mathbb {R} ^{2}$. We must show that there exists a $u=(u_{1},u_{2})^{T}\in \mathbb {R} ^{2}$ with $f(u)=v$. If we insert the definition of $f$, the vector $u$ we are looking for must thus satisfy $(u_{1}+\lambda u_{2},u_{2})^{T}=(v_{1},v_{2})^{T}$. That is, $u_{2}=v_{2}$ must hold. From this we get $u_{1}+\lambda v_{2}=v_{1}$, so $u_{1}=v_{1}-\lambda v_{2}$. If we set $u:=(v_{1}-\lambda v_{2},v_{2})^{T}\in \mathbb {R} ^{2}$, the following is true
$f(u)=f{\begin{pmatrix}v_{1}-\lambda v_{2}\\v_{2}\end{pmatrix}}={\begin{pmatrix}(v_{1}-\lambda v_{2})+\lambda v_{2}\\v_{2}\end{pmatrix}}={\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}=v.$
So $f$ is surjective.
Since $f$ is bijective and an endomorphism, it is also an automorphism.
Exercise (Non-commutativity in the endomorphism ring)
Let $V$ be an $n$-dimensional $K$-vector space with $n\geq 2$. Show: The endomorphism ring $\operatorname {End} _{K}(V)$ is not commutative.
Solution (Non-commutativity in the endomorphism ring)
Let $B=\{v_{1},\ldots ,v_{n}\}$ a basis of $V$, where by assumption $n\geq 2$ holds. We define two noncommutative endomorphisms $f,g\in \operatorname {Hom} _{K}(V)$ using the principle of linear continuation, by specifying the images of the basis vectors: For $k\in \{1,\ldots ,n\}$ set
$f:V\to V,\quad f(v_{k})={\begin{cases}v_{2}{\text{ if }}k=1,\\v_{1}{\text{ if }}k=2,\\v_{k}{\text{ else,}}\end{cases}}$
and
$g:V\to V,\quad g(v_{k})={\begin{cases}0{\text{ if }}k=1,\\v_{k}{\text{ else.}}\end{cases}}$
So the mapping $f$ swaps the first two basis vectors, while $g$ maps the first basis vector to the zero vector. In defining $f$ and $g$ we needed that there are $n\geq 2$ basis vectors. For $k\in \{1,\ldots ,n\}$ we now have
$(f\circ g)(v_{k})=f(g(v_{k}))={\begin{cases}0{\text{ if }}k=1,\\v_{1}{\text{ if }}k=2,\\v_{k}{\text{ else,}}\end{cases}}$
but
$(g\circ f)(v_{k})=g(f(v_{k}))={\begin{cases}v_{2}{\text{ if }}k=1,\\0{\text{ if }}k=2,\\v_{k}{\text{ else.}}\end{cases}}$
The basis vector $v_{2}$ is an element of the basis $B$ of $V$ , so it cannot be the zero vector. Therefore
$(g\circ f)(v_{1})=v_{2}\neq 0=(f\circ g)(v_{1}).$
So $g\circ f\neq f\circ g$ holds. Thus, the endomorphism ring $\operatorname {End} _{K}(V)$ is not commutative if $\dim(V)\geq 2$.
Exercise (Commutativity in the endomorphism ring)
Let $V$ be a one-dimensional $K$-vector space, i.e., $\dim(V)=1$. Show: that the endomorphism ring $\operatorname {End} _{K}(V)$ is commutative.
Solution (Commutativity in the endomorphism ring)
Let $B=\{b\}$ be a basis of $V$ and let $f,g\in \operatorname {End} _{K}(V)$ be arbitrary. Endomorphisms are already uniquely determined by the images of the basis vectors. Since there is only one basis vector due to $\dim(V)=1$, we have $f(b),g(b)\in \operatorname {span} (V)$ and thus
$f(b)=\lambda _{1}\cdot b\quad {\text{and}}\quad g(b)=\lambda _{2}\cdot b$
for certain $\lambda _{1},\lambda _{2}\in K$. Since $V=\operatorname {span} \{b\}$ , each $v\in V$ is of the form $v=\mu b$ for some $\mu \in K$. With linearity of $f$ and the commutativity of multiplication in $K$, it follows that
$f(v)=f(\mu b)=\mu f(b)=\mu (\lambda _{1}b)=(\mu \lambda _{1})b=(\lambda _{1}\mu )b=\lambda _{1}(\mu b)=\lambda _{1}v.$
Analogously one can show $g(v)=\lambda _{2}v$ for any $v\in V$ from which follows
$f\colon V\to V,\;v\mapsto \lambda _{1}\cdot v\quad {\text{and}}\quad g\colon V\to V,\;v\mapsto \lambda _{2}\cdot v.$
Thus, for all $v\in V$ we have
$(f\circ g)(v)=f(g(v))=f(\lambda _{2}v)=\lambda _{1}(\lambda _{2}v)=(\lambda _{1}\lambda _{2})v{\overset {(*)}{=}}(\lambda _{2}\lambda _{1})v=\lambda _{2}(\lambda _{1}v)=g(\lambda _{1}v)=g(f(v))=(g\circ f)(v),$
where in $(*)$ we have exploited the commutativity of multiplication in $K$.
Hint
In the above two problems we saw that $\operatorname {End} _{K}(V)$ is commutative if $\dim(V)=1$ and noncommutative if $\dim(V)\geq 2$. What is the situation for $\dim(V)=0$? If $\dim(V)=0$ , then $V$ is the null vector space. There is only one endomorphism over the null space $V=\{0\}$. This endomorphism is the null mapping $e\colon \{0\}\to \{0\},\ 0\mapsto 0$. So $\operatorname {End} _{K}(V)=\{e\}$ holds. Since the ring consists of only one element, $\operatorname {End} _{K}(V)$ is commutative.
Thus, $\operatorname {End} _{K}(V)$ is commutative if and only if $\dim(V)\leq 1$.
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