# Chapter 37. Automorphisms

# Chapter 37. Automorphisms

An automorphism is an isomorphism from a vector space to itself. If \(V\) is a vector space over a field \(F\), an automorphism of \(V\) is a linear map

$$
T:V\to V
$$

that is bijective. Equivalently, \(T\) is linear and has a linear inverse

$$
T^{-1}:V\to V.
$$

Thus an automorphism is a reversible linear transformation of one vector space. It changes the description or position of vectors inside the space, but it does not change the underlying linear structure. Standard references define an automorphism as an isomorphism from a vector space to itself.

## 37.1 Endomorphisms and Automorphisms

A linear map from a vector space to itself is called an endomorphism. Thus

$$
T:V\to V
$$

is an endomorphism when \(T\) is linear.

An automorphism is an invertible endomorphism. That is, \(T\) is an automorphism when there exists a linear map

$$
T^{-1}:V\to V
$$

such that

$$
T^{-1}\circ T=I_V
$$

and

$$
T\circ T^{-1}=I_V.
$$

The identity map

$$
I_V(v)=v
$$

is the simplest automorphism of \(V\). It changes nothing, and its inverse is itself.

Every automorphism is an endomorphism. Not every endomorphism is an automorphism. A projection, for example, is an endomorphism of \(\mathbb{R}^2\), but it loses information and has no inverse.

## 37.2 First Examples

Let

$$
V=\mathbb{R}^2.
$$

Define

$$
T
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
2x\\
3y
\end{bmatrix}.
$$

This map is linear. It is also invertible because

$$
T^{-1}
\begin{bmatrix}
u\\
v
\end{bmatrix} =
\begin{bmatrix}
u/2\\
v/3
\end{bmatrix}.
$$

Therefore \(T\) is an automorphism of \(\mathbb{R}^2\).

Now define

$$
P
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
x\\
0
\end{bmatrix}.
$$

This map is linear, but it is not an automorphism. It sends every vector on the \(y\)-axis to zero:

$$
P
\begin{bmatrix}
0\\
y
\end{bmatrix} =
\begin{bmatrix}
0\\
0
\end{bmatrix}.
$$

Thus \(P\) is not injective. It collapses the plane onto a line, so it cannot be reversed.

## 37.3 Automorphisms of Coordinate Space

An automorphism of \(F^n\) is exactly an invertible linear map

$$
T:F^n\to F^n.
$$

Every such map has the form

$$
T(x)=Ax
$$

for some \(n\times n\) matrix \(A\). The map \(T\) is an automorphism exactly when \(A\) is invertible.

Thus automorphisms of \(F^n\) correspond to invertible \(n\times n\) matrices.

The set of all invertible \(n\times n\) matrices over \(F\) is called the general linear group and is denoted

$$
GL_n(F).
$$

Therefore,

$$
\operatorname{Aut}(F^n)\cong GL_n(F).
$$

Here \(\operatorname{Aut}(F^n)\) denotes the set of all automorphisms of \(F^n\).

## 37.4 The Automorphism Group

The automorphisms of a vector space form a group under composition.

Let

$$
\operatorname{Aut}(V)
$$

denote the set of all automorphisms of \(V\).

This set has four group properties.

First, the composition of two automorphisms is an automorphism. If

$$
S,T\in \operatorname{Aut}(V),
$$

then

$$
S\circ T
$$

is linear and invertible. Its inverse is

$$
(S\circ T)^{-1}=T^{-1}\circ S^{-1}.
$$

Second, composition is associative because function composition is associative:

$$
R\circ(S\circ T)=(R\circ S)\circ T.
$$

Third, the identity transformation \(I_V\) is an automorphism and satisfies

$$
I_V\circ T=T
$$

and

$$
T\circ I_V=T.
$$

Fourth, every automorphism \(T\) has an inverse \(T^{-1}\), and this inverse is also an automorphism.

Thus \(\operatorname{Aut}(V)\) is a group.

## 37.5 Matrix Form of the Automorphism Group

If \(V\) is finite-dimensional and a basis \(B\) is chosen, each automorphism \(T:V\to V\) has a matrix

$$
[T]_B.
$$

Since \(T\) is invertible, the matrix \([T]_B\) is invertible. Conversely, every invertible matrix in this basis defines an automorphism of \(V\).

Thus, after choosing a basis,

$$
\operatorname{Aut}(V)
$$

is represented by

$$
GL_n(F),
$$

where

$$
n=\dim(V).
$$

The phrase “after choosing a basis” is important. The automorphisms are intrinsic maps \(V\to V\). The matrices are coordinate descriptions of those maps.

If a different basis is chosen, the same automorphism is represented by a similar matrix.

## 37.6 Characterizations

Let \(V\) be finite-dimensional, and let

$$
T:V\to V
$$

be linear. The following conditions are equivalent:

| Condition | Meaning |
|---|---|
| \(T\) is an automorphism | \(T\) is invertible |
| \(T\) is injective | No nonzero vector is collapsed |
| \(T\) is surjective | Every vector is reached |
| \(\ker(T)=\{0\}\) | The kernel is trivial |
| \(\operatorname{im}(T)=V\) | The image is the whole space |
| \(\operatorname{rank}(T)=\dim(V)\) | Full rank |
| \([T]_B\) is invertible | Any basis matrix is invertible |
| \(\det([T]_B)\neq 0\) | The determinant is nonzero |

These equivalences are special to maps from a finite-dimensional vector space to itself. In this setting, injectivity and surjectivity imply each other by rank-nullity.

## 37.7 Proof Using Rank-Nullity

Let

$$
T:V\to V
$$

be linear, where \(V\) is finite-dimensional.

The rank-nullity theorem gives

$$
\dim(V)=\operatorname{rank}(T)+\operatorname{nullity}(T).
$$

If \(T\) is injective, then

$$
\ker(T)=\{0\},
$$

so

$$
\operatorname{nullity}(T)=0.
$$

Hence

$$
\operatorname{rank}(T)=\dim(V).
$$

Since the image is a subspace of \(V\) with the same dimension as \(V\), we have

$$
\operatorname{im}(T)=V.
$$

Thus \(T\) is surjective.

Conversely, if \(T\) is surjective, then

$$
\operatorname{rank}(T)=\dim(V).
$$

Rank-nullity gives

$$
\operatorname{nullity}(T)=0.
$$

Therefore

$$
\ker(T)=\{0\},
$$

so \(T\) is injective.

Thus, for endomorphisms of finite-dimensional spaces, either injectivity or surjectivity is enough to prove that the map is an automorphism.

## 37.8 Automorphisms Preserve Bases

An automorphism sends bases to bases.

Let

$$
T:V\to V
$$

be an automorphism, and let

$$
B=(v_1,\ldots,v_n)
$$

be a basis of \(V\).

We claim that

$$
T(B)=(T(v_1),\ldots,T(v_n))
$$

is also a basis of \(V\).

First, it is linearly independent. Suppose

$$
c_1T(v_1)+\cdots+c_nT(v_n)=0.
$$

By linearity,

$$
T(c_1v_1+\cdots+c_nv_n)=0.
$$

Since \(T\) is injective,

$$
c_1v_1+\cdots+c_nv_n=0.
$$

Since \(B\) is linearly independent,

$$
c_1=\cdots=c_n=0.
$$

Second, it spans \(V\). Let \(w\in V\). Since \(T\) is surjective, there exists \(v\in V\) such that

$$
T(v)=w.
$$

Write

$$
v=c_1v_1+\cdots+c_nv_n.
$$

Then

$$
w=T(v)=c_1T(v_1)+\cdots+c_nT(v_n).
$$

Thus \(T(B)\) spans \(V\).

Therefore \(T(B)\) is a basis.

## 37.9 Automorphisms and Change of Basis

Every ordered basis of \(V\) can be obtained from any other ordered basis by a unique automorphism.

Let

$$
B=(v_1,\ldots,v_n)
$$

and

$$
C=(w_1,\ldots,w_n)
$$

be ordered bases of \(V\).

Define

$$
T(v_i)=w_i
$$

for each \(i\), and extend linearly:

$$
T(c_1v_1+\cdots+c_nv_n) =
c_1w_1+\cdots+c_nw_n.
$$

This map is linear. Since \(C\) is a basis, \(T\) is bijective. Hence \(T\) is an automorphism.

It is unique because a linear map is determined by its values on a basis.

Thus automorphisms describe changes from one basis to another. They are the reversible linear changes of coordinate frame.

## 37.10 Similarity and Automorphisms

Let

$$
A:V\to V
$$

be a linear operator, and let

$$
P:V\to V
$$

be an automorphism.

The operator

$$
P^{-1}AP
$$

is the same operator \(A\) viewed through the coordinate change \(P\). In matrix language, this is a similarity transformation.

If \(A\) has matrix \(M\) in one basis, and \(P\) is the change-of-basis matrix, then the matrix in the new basis has the form

$$
P^{-1}MP.
$$

Similar matrices represent the same linear operator under different bases.

Automorphisms are therefore the transformations that change coordinates without changing the vector-space structure.

## 37.11 Inner Automorphisms of the Matrix Algebra

Let

$$
P\in GL_n(F).
$$

The rule

$$
A\mapsto P^{-1}AP
$$

maps \(n\times n\) matrices to \(n\times n\) matrices.

This map is an automorphism of the algebra of matrices: it is linear, invertible, and preserves multiplication.

It is linear because

$$
P^{-1}(A+B)P=P^{-1}AP+P^{-1}BP
$$

and

$$
P^{-1}(cA)P=cP^{-1}AP.
$$

It preserves products because

$$
(P^{-1}AP)(P^{-1}BP)=P^{-1}A(PP^{-1})BP=P^{-1}ABP.
$$

It is invertible, with inverse

$$
A\mapsto PAP^{-1}.
$$

This construction is called conjugation by \(P\). It is central in the study of canonical forms.

## 37.12 Fixed Vectors

An automorphism may leave some vectors fixed.

A vector \(v\in V\) is fixed by \(T\) if

$$
T(v)=v.
$$

The set of fixed vectors is

$$
\operatorname{Fix}(T)=\{v\in V:T(v)=v\}.
$$

This is the kernel of \(T-I\):

$$
\operatorname{Fix}(T)=\ker(T-I).
$$

Therefore \(\operatorname{Fix}(T)\) is a subspace of \(V\).

For example, a reflection of \(\mathbb{R}^2\) across a line fixes every vector on that line and reverses the perpendicular direction. The fixed subspace is the line of reflection.

## 37.13 Periodic Automorphisms

An automorphism \(T\) is periodic if

$$
T^k=I
$$

for some positive integer \(k\).

For example, a rotation of the plane by \(90^\circ\) satisfies

$$
T^4=I.
$$

After four applications, every vector returns to its original position.

In matrix form, a periodic automorphism is represented by an invertible matrix \(A\) satisfying

$$
A^k=I.
$$

Such transformations are important because their powers form a finite cyclic subgroup of \(\operatorname{Aut}(V)\).

## 37.14 Automorphisms of One-Dimensional Spaces

Let \(V\) be a one-dimensional vector space over \(F\). Choose a nonzero vector \(v\) as a basis.

Every vector in \(V\) has the form

$$
cv.
$$

A linear map \(T:V\to V\) is determined by \(T(v)\). If \(T\) is an automorphism, then \(T(v)\neq 0\). Therefore

$$
T(v)=av
$$

for some nonzero scalar \(a\in F\).

Thus every automorphism of a one-dimensional vector space is multiplication by a nonzero scalar.

So

$$
\operatorname{Aut}(V)\cong F^\times,
$$

where \(F^\times\) is the multiplicative group of nonzero elements of \(F\).

## 37.15 Automorphisms of \(\mathbb{R}^2\)

Automorphisms of \(\mathbb{R}^2\) are exactly invertible \(2\times 2\) real matrices.

Let

$$
A=
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}.
$$

The map

$$
T(x)=Ax
$$

is an automorphism exactly when

$$
\det(A)=ad-bc\neq 0.
$$

Geometrically, such a map sends the plane to itself without collapsing area to zero. It may rotate, reflect, shear, stretch, or combine these operations.

If

$$
\det(A)>0,
$$

the transformation preserves orientation.

If

$$
\det(A)<0,
$$

the transformation reverses orientation.

If

$$
\det(A)=0,
$$

the transformation collapses the plane into a line or a point, and it is not an automorphism.

## 37.16 Automorphisms and Determinants

For finite-dimensional spaces, determinants give a scalar test for automorphisms.

Let \(T:V\to V\), and choose a basis \(B\). Let

$$
A=[T]_B.
$$

Then

$$
T \text{ is an automorphism}
\quad\Longleftrightarrow\quad
\det(A)\neq 0.
$$

Although the matrix \(A\) depends on the basis, the condition \(\det(A)\neq 0\) does not.

If \(C\) is another basis, the matrix of \(T\) in basis \(C\) is similar to \(A\):

$$
A_C=P^{-1}AP.
$$

Then

$$
\det(A_C)=\det(P^{-1}AP).
$$

Using multiplicativity of determinant,

$$
\det(A_C)=\det(P^{-1})\det(A)\det(P)=\det(A).
$$

So the determinant of an operator is basis-independent.

## 37.17 Automorphisms and Eigenvalues

Let

$$
T:V\to V
$$

be a linear operator on a finite-dimensional vector space.

If \(T\) has eigenvalue \(0\), then there exists a nonzero vector \(v\) such that

$$
T(v)=0v=0.
$$

Thus

$$
v\in\ker(T),
$$

so \(T\) is not injective and cannot be an automorphism.

Conversely, if \(T\) is not an automorphism, then

$$
\ker(T)\neq \{0\}
$$

for finite-dimensional \(V\). Hence there is a nonzero \(v\) such that

$$
T(v)=0.
$$

Thus \(0\) is an eigenvalue.

Therefore,

$$
T \text{ is an automorphism}
\quad\Longleftrightarrow\quad
0 \text{ is not an eigenvalue of } T.
$$

For a matrix \(A\), this is equivalent to saying that its characteristic polynomial does not vanish at \(0\), or equivalently that

$$
\det(A)\neq 0.
$$

## 37.18 Automorphisms of Polynomial Spaces

Let \(P_2\) be the vector space of polynomials of degree at most \(2\). Define

$$
T:P_2\to P_2
$$

by

$$
T(p)(x)=p(x+1).
$$

This is linear because

$$
T(p+q)(x)=(p+q)(x+1)=p(x+1)+q(x+1)
$$

and

$$
T(cp)(x)=cp(x+1)=cT(p)(x).
$$

It is invertible. The inverse is

$$
T^{-1}(p)(x)=p(x-1).
$$

Therefore \(T\) is an automorphism of \(P_2\).

Using the basis

$$
(1,x,x^2),
$$

compute:

$$
T(1)=1,
$$

$$
T(x)=x+1,
$$

$$
T(x^2)=(x+1)^2=x^2+2x+1.
$$

Thus the matrix of \(T\) in this basis is

$$
[T]=
\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}.
$$

This matrix is upper triangular with diagonal entries all equal to \(1\). Hence its determinant is \(1\), so it is invertible.

## 37.19 Nonexamples

The zero map

$$
Z(v)=0
$$

is an endomorphism of \(V\), but it is an automorphism only when \(V=\{0\}\). If \(V\) has any nonzero vector, the zero map sends that vector to zero and is not injective.

A projection

$$
P^2=P
$$

is usually not an automorphism. If \(P\neq I\), then it collapses some directions or fails to reach all of \(V\).

A nilpotent map

$$
N^k=0
$$

cannot be an automorphism unless \(V=\{0\}\). If \(N\) were invertible, then \(N^k\) would also be invertible. But the zero map is not invertible on a nonzero space.

A map with determinant zero is not an automorphism. Its matrix loses rank.

## 37.20 Automorphisms as Symmetries

An automorphism may be viewed as a symmetry of the vector-space structure.

It preserves all linear facts: zero, sums, scalar multiples, linear combinations, subspaces, bases, dimension, kernels, images, rank, and linear independence.

It may fail to preserve extra structure unless the automorphism is required to respect that structure. For example, a general automorphism of \(\mathbb{R}^n\) need not preserve lengths or angles. To preserve those, one studies orthogonal transformations.

Thus there are different levels of symmetry:

| Structure preserved | Transformations |
|---|---|
| Vector addition and scalar multiplication | Automorphisms |
| Lengths and angles | Orthogonal or unitary transformations |
| Orientation and volume form | Special linear transformations |
| Inner product and orientation | Special orthogonal transformations |

Automorphisms are the broadest reversible linear symmetries of a vector space.

## 37.21 Summary

An automorphism is an isomorphism from a vector space to itself.

For a vector space \(V\), an automorphism is a linear map

$$
T:V\to V
$$

that is invertible. The set of all automorphisms is denoted

$$
\operatorname{Aut}(V).
$$

Under composition, \(\operatorname{Aut}(V)\) forms a group.

If \(V\) is finite-dimensional and

$$
\dim(V)=n,
$$

then, after choosing a basis,

$$
\operatorname{Aut}(V)
$$

is represented by the general linear group

$$
GL_n(F).
$$

A linear endomorphism \(T:V\to V\) is an automorphism exactly when any of the following equivalent conditions holds:

$$
\ker(T)=\{0\},
$$

$$
\operatorname{im}(T)=V,
$$

$$
\operatorname{rank}(T)=\dim(V),
$$

$$
\det([T]_B)\neq 0.
$$

Automorphisms are reversible linear transformations. They express the internal symmetries of a vector space and provide the algebraic language for change of basis, similarity, and coordinate transformations.