1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook