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Regarding the Form of an Automorphism

  1. Dec 14, 2007 #1
    [SOLVED] Regarding the Form of an Automorphism

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

    [itex]f : \mathbb{R}^n \rightarrow \mathbb{R}^n[/itex] is an automorphism if and only if [itex]f[/itex] has the form

    [tex]\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}[/tex]


    [tex]g_i(x_1, \ldots, x_n) = a_{i1}x_1 + \ldots + a_{in}x_n[/tex]

    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. jcsd
  3. Dec 14, 2007 #2


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    Try using the standard basis vectors for R^n.
  4. Dec 16, 2007 #3
    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
  5. Dec 16, 2007 #4


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    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.
  6. Dec 17, 2007 #5
    Ah, I see now. Man, I feel stupid for not being able to see that. Thanks a lot.
  7. Dec 21, 2007 #6
    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:


    [tex]\vec{x} = \begin{pmatrix}x_1 \\ \hdots \\ x_n\end{pmatrix}[/tex]


    [tex]\vec{y} = \begin{pmatrix}y_1 \\ \hdots \\ y_n\end{pmatrix}[/tex]

    Let a and b be two scalars.

    [tex]\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*}[/tex]

    Note that I've employed the fact that [itex]g_i[/itex] is a homomorphism.

    However, showing that f is bijective is hard:

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

    Another thing to notice is that [itex]f(\vec{x} - \vec{y}) = \vec{0}[/itex] implies that

    [tex]a_{i1}(x_1 - y_1) + \ldots + a_{in}(x_n - y_n) = 0[/tex]

    for i = 1, ..., n. This yields a homogeneous system of linear equations. Obviously, [itex]x_j = y_j[/itex] 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 [itex]\vec{y}[/itex] in the codomain, what vector [itex]\vec{x}[/itex] satisfies [itex]\vec{y} = f(\vec{x})[/itex]? I can write [itex]\vec{y}[/itex] 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.
  8. Dec 22, 2007 #7


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    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
  9. Dec 22, 2007 #8
    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?
  10. Dec 22, 2007 #9
    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
  11. Dec 22, 2007 #10
    Dang it. Why didn't I think of that. If f is given by

    [tex]\binom{x}{y} \mapsto \binom{x}{0}[/tex]

    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.
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