Results 1 to 6 of 6

Math Help - Homomorphism

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    18

    Homomorphism



    So I want to show f(g1 + g2) = f(g1) + f(g2)?

    I'm confused as to what exactly it is i'm adding, I would guess a generall 2x2 matrix i.e ((a b c d) + (e f g h)) Is this right? I don't understand exactly where O2 comes into it...

    I know it's not an Isomorphism as it's not injective, due to a rotation of 2*pi, which means it's not bijective and thus not an isomorphism.

    I'd guess that reflection on the other hand is an isomorphism as no single reflection gives itself, i.e you'd have to reflect through the same line twice.

    Is this corect? Help would be appreciated mainly for the first part though.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Hi

    Take care! The law for matrices in O_2 will be multiplication (O_2\subset GL(2,\mathbb{R}). Addition can't be used: I_2\in O_2 but 2I_2\notin O_2 (since its determinant is 4)
    Do you now see why \theta\mapsto rot_{\theta} is a homomorphism?

    About isomorphisms, you're right for rotations.

    But before thinking wether the second function is an isomorphism or not, is it a homomorphism?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    18
    Not sure really. Do I want to show that two matrices added together, then apply the function f, but i don't know where to apply the fact it's in I_2...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    We have rot_{\theta}=\begin{pmatrix}cos\theta & -sin\theta \\ sin\theta & cos\theta\end{pmatrix}

    To prove that f is a morphism, what you have to show is that, if a,b are two reals, rot_{a+b}=:f(a+b)=f(a)f(b):=rot_arot_b, i.e.

    \begin{pmatrix}cos(a+b) & -sin(a+b) \\ sin(a+b) & cos(a+b)\end{pmatrix}=\begin{pmatrix}cosa & -sina \\ sina & cosa\end{pmatrix}\begin{pmatrix}cosb & -sinb \\ sinb & cosb\end{pmatrix}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2009
    Posts
    18
    Thanks for the help so far, so now i've expanded the cos(a+b) and sin(a+b) into the two relevant trig identities, but I can't see how to get it into the final form without cheating as such. Is there a factor i can take out?

    Also i've managed to prove that refθ is not a HM, I did this by using "Z's" do you reckon this is ok, or do I have to use the reflection matrix again of (cosθ sinθ sinθ -cosθ)?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    The easiest way I think to prove the matrix equality is to compute the product and check that the matrix you obtain is the one you got by expanding cos(a+b) and sin(a+b).

    To prove that ref is not a homomorphism, you can show that the composition of two reflections over a line in \mathbb{R}^2 is not always a reflection over a line, hence you just need one example. So there are a lot of good answers, but I'm sorry I don't understand "Z's" ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 19th 2013, 04:05 PM
  2. Homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 15
    Last Post: June 25th 2011, 07:45 AM
  3. Is this a homomorphism?
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: August 27th 2010, 02:10 AM
  4. homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 21st 2009, 10:38 PM
  5. Homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 15th 2009, 03:17 PM

Search Tags


/mathhelpforum @mathhelpforum