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Linear map is isomorphic

  1. Oct 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Let V={a cosx + b sinx | a,b [tex]\in[/tex] R}

    (a) Show that V is a subspace of the R-vector space of all maps from R to R.

    (b) Show that V is isomorphic to R^2, under the map
    f: V[tex]\rightarrow[/tex]R^2
    a cosx + b sinx [tex]\rightleftharpoons[/tex] [ a over b ] (this is supposed to be a matrix with a above and b below, couldn't find it in the Latex reference)



    3. The attempt at a solution

    I have done part (a) okay so it's just part (b) I need a hand with. So I know V is isomorphic to R^2 if the map f is linear and the dimV = dim R^2.

    I think I sort of showed that the dimensions are equal by taking a basis of {0 , cosx + sinx}. Is that basis okay? And am I right in thinking that because that has two elements dimV=2 and obviously dimR^2= 2 yeah?

    So assuming that all that's ok so far I'm kind of stuck showing that it's linear. The properties of a linear map are f(u+v)=f(u)+f(v) and f(av)= af(v) where u,v[tex]\in[/tex]V and a[tex]\in[/tex] R. I'm trying to show those properties are true at the minute but having some difficulty. Am I on the right track at least?

    thanks in advance!
     
  2. jcsd
  3. Oct 25, 2010 #2

    Office_Shredder

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    How do you construct sinx from the set {0, cosx + sinx}?

    Also any set containing zero is not linearly independent since 1*0=0 so there is a non-trivial linear combination which gives zero.
     
  4. Oct 25, 2010 #3

    HallsofIvy

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    No, it is NOT okay. A basis never includes the 0 vector. Try {cos x, sin x} instead.
    "\begin{bmatrix} a \\ b\end{bmatrix}" gives
    [tex]\begin{bmatrix} a \\ b\end{bmatrix}[/tex]

    If u= a cos(x)+ b sin(x) and v= c cos(x)+ d sin(x), what is u+ v?

    [tex]f(u)= \begin{bmatrix}a \\ b\end{bmatrix}[/tex]
    and
    [tex]f(v)= \begin{bmatrix}c \\ d \end{bmatrix}[/tex]

    what is f(u+ v)?
     
    Last edited by a moderator: Oct 25, 2010
  5. Oct 25, 2010 #4
    Thanks for that. Looking back on it the basis I picked is obviously not a basis I just wasn't thinking. Spelling out for me what f(u) equals really helped. When I came back to it I solved it in about 10mins!

    Thanks again! :)
     
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