1. Not finding help here? Sign up for a free 30min 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!

Simple isomorphism question

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Is the following transformation an isomorphism:

    [itex]a_0+bx+cx^{2}+dx^{3} \rightarrow \begin{bmatrix} a & b\\ c & d \end{bmatrix}[/itex]

    2. Relevant equations

    A transformation is an isomorphism if:

    1. The transformation is one-to-one
    2. The transformation is onto



    3. The attempt at a solution

    I took an accelerated Linear Algebra course over the summer, and in the last lecture my professor barely touched the concepts of "onto" and "one-to-one"

    I know a transformation is onto if T:V→W it maps to every vector in W (essentially no restrictions)
    and that a transformation is one to one if for each vector in V maps to only one vector in W.

    I have working through a bunch of problems by finding the transformation matrix and then taking its determinant to see if its one to one (i.e if det(T) ≠ 0) or finding the kernel (if Ker(T)=0)
    I was testing if the transformation was onto by finding RREF and seeing if any restrictions popped up (if not then it was onto, I think this was right)

    This problem seems extremely simple but I'm not sure where to start since I don't know what T looks like and haven't had much guidance in this type of problem yet (none actually)

    Any help is appreciated, thanks!
     
    Last edited: Aug 5, 2014
  2. jcsd
  3. Aug 5, 2014 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You need to define this transformation more carefully. What exactly is its domain?

    A linear transformation is an isomorphism (to be more specific, a vector space isomorphism) if 1 and 2 hold. So if V and W are vector spaces, and you suspect that a function ##T:V\to W## may be an isomorphism, you should usually start by verifying that it's linear.

    Your descriptions of these terms are accurate, but I will still show you the definitions I use.

    The terms "injective" and "surjective" are more popular. A function ##f:A\to B## is injective if the implication ##f(x)=f(y)\Rightarrow x=y## holds for all ##x,y\in A##, and surjective if ##f(A)=B##, i.e. if the range is equal to the codomain (rather than a proper subset of it). So a proof of injectivity should start with "Let x and y be arbitrary elements of A such that f(x)=f(y)". Then you prove that x=y. A proof of surjectivity should usually start with "let y be an arbitrary element of B". Then you find an x in A such that f(x)=y.
     
  4. Aug 5, 2014 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Let's call ##P_3## the linear space of 3rd degree polynomials and ##M## the 2 by 2 matrices. You don't need any representation other than what you are given to see if the given transformation, call it ##T##, is 1-1 and onto.

    If ##T(p_1) = T(p_2)## can you show ##p_1 = p_2## using the usual properties of polynomials and matrices? Similarly, if ##A\in M## can you find ##p\in P_3## such that ##T(p)= A##?
     
  5. Aug 5, 2014 #4
    T:P3→M2,2


    Ok,
    I do know that its linear if: k(Tu)=T(ku) and T(u+v)=T(u)+T(v), thats simple enough.



    Nice,
    I like these definitions better than the ones I was working with.
     
  6. Aug 5, 2014 #5

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right, so you should show that for all ##u,v\in P_3## and all ##a,b\in\mathbb R##, we have ##T(au+bv)=aTu+bTv##. Or you can break it up in two separate statements, as you did here:

     
  7. Aug 5, 2014 #6
    I have the tools to but im not really sure where to start.
    " If ##T(p_1) = T(p_2)## can you show ##p_1 = p_2## using the usual properties of polynomials and matrices?" I know this is the test for 1-1

    and

    " Similarly, if ##A\in M## can you find ##p\in P_3## such that ##T(p)= A##?" is the test for onto, but im not sure where to start. What properties of polynomials/matrices to I use?
     
  8. Aug 5, 2014 #7
    Right, and it turns out T is linear. Since the two statements hold. I'm just confused on how to show 1-1 and onto. Before in other problems, I could find T (or was given T). Now, since I don't know T, im a little confused on how to test its "onto-ness" or "one to one-ness".
     
  9. Aug 5, 2014 #8
    This is what I mean:
    A simple problem that popped up with the same question was:

    [itex]c_{0}+c_{1}x\rightarrow (c_{0}-c_{1}, c_{1})
    [/itex]

    with T:P1→R2

    This case is simple since I can find T, its just \begin{bmatrix} 1 &-1 \\ 0& 1 \end{bmatrix}
    I could easily find its det (test for 1-1) and RREF restrictions (onto test)

    On this problem, it just seems more vague....
     
  10. Aug 5, 2014 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are making this much harder than it is. Say you are given$$
    A = \begin{bmatrix}a & b \\ c & d\end{bmatrix} \in M$$Can you write down a polynomial in ##P_3## that maps to it? That would show ##T## is onto.
     
  11. Aug 5, 2014 #10
    sure, wouldnt It just be: [itex]a_{0}+bx+cx^{2}+dx^{3}[/itex]?

    edit: ahh...so that it? If the polynomial given was say: [itex]a_{0}+bx+cx^{2}+d(x+1)^{3}[/itex] then it wouldnt?
     
  12. Aug 5, 2014 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Simple, isn't it. Now show ##T## is 1-1.
     
  13. Aug 5, 2014 #12
    I got it from here, Thanks LCKurtz and Fredrik.Im making it way harder than it it. Ohh well, now I know!

    Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple isomorphism question
  1. Isomorphism question (Replies: 2)

  2. Isomorphism question (Replies: 0)

Loading...