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Proving a linear transformation is an isomorphism

  1. Dec 2, 2009 #1
    Define T: F^2 --> P_1(F) by T(a, b) = a + bx (with P_1 denoting P sub 1)

    I usually prove problems such as this by constructing a matrix of T using bases for the vector spaces and then proving that the matrix is invertible, but is the following also a viable proof that T is an isomorphism? I know it is not finished, but is it a step in the right direction?

    let T(z) = m + nx (T(z) is contained in P_1(F))

    so z = (m, n) (z is an element of F^2)

    this means that that the general form for all elements in P_1(F) has a pre-image in F^2, which means that T is onto(?), so therefore T is invertible and F^2 is isomorphic to P_1(F).

    Is this any start at all? Any suggestions?
  2. jcsd
  3. Dec 3, 2009 #2


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    To prove a function is an isomorphism, you must prove:
    1) That it "respects" all operations: that f(a*b)= f(a)*f(b) or any operation *.
    2) That is it one-to-one.
    3) That it is onto.

    Here, the operations are vector addition and scalar multiplication.
    T((a,b)+ (c,d))= T((a+c,b+d))= (a+c)+ (b+d)x= a+ bx+ c+ dx= T((a,b))+ T(c,d)
    T(\alpha (a, b))= T(\alpha a, \alpha b)= \alpha a+ (\alpha b)x= \alpha(a+ bx)= \alpha T((a,b))

    To show "one-to-one", suppose (a,b) and (c,d) were such that T((a,b))= T((c,d)). Then a+ bx= c+ dx for all x[/itex]. What follows from that? (Take x= 0 or x= 1, for example.)

    To show "onto", given the linear polynomial u+ vx, does there exist (a,b) such that T((a,b))= u+ vx?
  4. Dec 3, 2009 #3
    So in this case you use addition and multiplication, but is that because those are the operations used in the transformation, or because you are simply proving it is a linear transformation?

    As for showing that T is one-to-one, is this the right idea?

    T((a,b)) = T((c,d))
    so then a + bx = c + dx.
    then letting x = 0, a = c.
    when x = 1, a + b = c + d, and since a = c, we see that b = d.
    so (a,b) = (c,d). this means that T is one-to-one.
    is this correct?
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