# Regarding the Form of an Automorphism

1. Dec 14, 2007

### e(ho0n3

[SOLVED] Regarding the Form of an Automorphism

The problem statement, all variables and given/known data
I want to prove the following:

$f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is an automorphism if and only if $f$ has the form

$$\begin{pmatrix}x_1 \\ \vdots \\ x_n \end{pmatrix} \mapsto \begin{pmatrix} g_1(x_1, \ldots, x_n) \\ \vdots \\ g_n(x_1, \ldots, x_n) \end{pmatrix}$$

where

$$g_i(x_1, \ldots, x_n) = a_{i1}x_1 + \ldots + a_{in}x_n$$

and the a's are constants.

The attempt at a solution
Proving that f is an automorphism given that it has the latter form is easy. Proving the converse is hard.

It took me a while to prove the case for n = 1: For x not equal to 0, f(x)/x = f(x/x) = f(1) so f(x) = f(1)x. For x = 0, f(0) = f(0*0) = 0f(0) = f(1)0.

I was thinking about the 'n = 2' case. I can't employ the same method of proof as before since I'm dealing with vectors now. I'm thinking that the preservation of structure, i.e. f(u + v) = f(u) + f(v) and f(ru) = rf(u), must in some fashion force f to have the form described in the statement. However, I don't see how this could possibly be.

2. Dec 14, 2007

### morphism

Try using the standard basis vectors for R^n.

3. Dec 16, 2007

### e(ho0n3

In what way? I thought about u and f(u) as linear combinations of the basis vectors and the structure preserving properties. I didn't obtain further insight into the problem.

Perhaps another hint is in order.

Last edited: Dec 17, 2007
4. Dec 16, 2007

### morphism

Let {e1,...,en} be the standard basis of R^n, and let u1=f(e1),...,un=f(en). Then given x in R^n, with x=x1e1+...+xnen, we have f(x)=x1u1+...+xnun.

5. Dec 17, 2007

### e(ho0n3

Ah, I see now. Man, I feel stupid for not being able to see that. Thanks a lot.

6. Dec 21, 2007

### e(ho0n3

I wrote that thinking that it would be easy but as it turns out, it isn't. More specifically, showing that f preserves structure is easy:

Let

$$\vec{x} = \begin{pmatrix}x_1 \\ \hdots \\ x_n\end{pmatrix}$$

and

$$\vec{y} = \begin{pmatrix}y_1 \\ \hdots \\ y_n\end{pmatrix}$$

Let a and b be two scalars.

\begin{align*}f(a\vec{x} + b\vec{y}) &= f\left(\begin{pmatrix}ax_1 + by_1 \\ \hdots \\ ax_n + by_n\end{pmatrix}\right) = \begin{pmatrix} g_1(ax_1 + by_1, \ldots, ax_n +by_n) \\ \vdots \\ g_n(ax_1 + by_1, \ldots, ax_n + by_n) \end{pmatrix} = \begin{pmatrix} ag_1(x_1, \ldots, x_n) + bg_1(y_1, \ldots, y_n) \\ \vdots \\ ag_n(x_1, \ldots, x_n) + bg_n(y_1, \ldots, y_n) \end{pmatrix}\\ &= a\begin{pmatrix} g_1(x_1, \ldots, x_n) \\ \vdots \\ g_n(x_1, \ldots, x_n) \end{pmatrix} + b\begin{pmatrix} g_1(y_1, \ldots, yx_n) \\ \vdots \\ g_n(y_1, \ldots, y_n) \end{pmatrix} = af(\vec{x}) + bf(\vec{y})\end{align*}

Note that I've employed the fact that $g_i$ is a homomorphism.

However, showing that f is bijective is hard:

To demonstrate that f is injective, suppose $f(\vec{x}) = f(\vec{y})$. Then $f(\vec{x}) - f(\vec{y}) = \vec{0}$ which by the above implies that $f(\vec{x} - \vec{y}) = \vec{0}$. $\vec{x} - \vec{y}$ could very well be $\vec{0}$ since $f(\vec{0}) = \vec{0}$. However, there may be another vector $\vec{v} \ne \vec{0}$ such that $f(\vec{v}) = \vec{0}$.

Another thing to notice is that $f(\vec{x} - \vec{y}) = \vec{0}$ implies that

$$a_{i1}(x_1 - y_1) + \ldots + a_{in}(x_n - y_n) = 0$$

for i = 1, ..., n. This yields a homogeneous system of linear equations. Obviously, $x_j = y_j$ is one solution. I don't know how to show that this is the only solution though.

Showing that f is surjective is just as difficult: Given a vector $\vec{y}$ in the codomain, what vector $\vec{x}$ satisfies $\vec{y} = f(\vec{x})$? I can write $\vec{y}$ as a linear combination of the natural basis vectors, but then I would need to determine which vectors the basis vectors are the image of.

7. Dec 22, 2007

### morphism

I wanted to comment on this earlier, but I assumed you just forgot to post that there were some conditions on the g_i. As it stands, the definition of f you're given does not guarantee that it will be an automorphism. For example if g_1 = g_2 = ... = g_n = 0, then certainly the g_i satisfy the requirements, and f(x)=0 for all x.

Last edited: Dec 22, 2007
8. Dec 22, 2007

### e(ho0n3

You're right.

I'm trying to think of a non-trivial map that is a linear transformation but is not an automorphism. So far I can't think of any. Perhaps all non-trivial linear transformations are automorphisms?

9. Dec 22, 2007

### andytoh

Any non-injective or non-surjective endomorphism is not an automorphism. For example, any projection onto a nontrivial proper subspace, e.g. project R^2 onto the x-axis.

Last edited: Dec 22, 2007
10. Dec 22, 2007

### e(ho0n3

Dang it. Why didn't I think of that. If f is given by

$$\binom{x}{y} \mapsto \binom{x}{0}$$

then f is a non-trivial linear transformation that is not injective and hence not an automorphism.

I will mark this thread as solved. Thanks everyone.