# Proving a linear transformation is an isomorphism

1. Dec 2, 2009

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. Dec 3, 2009

### HallsofIvy

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?

3. Dec 3, 2009