One to One and Onto Transformation

Ok, it's been a while since I cracked an abstract algebra book, but here goes.

1) to show that it is one-to-one you have to show that for each input there is an output. Your definition says we are going from P_2 to P_2. I don't know what set P_2 is, but you have to show that for every input from P_2 there is an output in P_2.

2) to show that it is onto, you have to show that for every possible output in P_2, there is a member of P_2 that yields it when used as an input.

3) is T^-1 indicating the inverse? If so, I'll get back to you on that one.

FreeDuck, P_2 is the space of all polynomials of degree less than or equal to 2 and T^-1 indicated the inverse transformation. Thanks for you time.

For one-to-one, you have to show that each input maps to exactly one output.

For the inverse, think of a function that maps (2b-a, b+a) to (a, b).

Hints:What happens when you add 2b-a to b+a? And what happens when you subtract 2b-a from 2*(b+a)?

RK4

Ah first year linear algebra and isomorphic transitions!

As markr said for 1-1, a proof by contradiction would be easy. Assume its not 1-1, show two distinct points X and X+e (for e <>0) both map to the same value (line). And then simply and reduce both terms to show this can only happen if e=0, contradicting your intital supposition that it is not 1-1! This should be at most a 5 line proof or less!

For onto you could do a similar proof by contradiction, then show its false.

Hint on the inverse

T^-1(T(ax + b)) = ax + b, or

T^-1(( 2b - a )*x + ( b+a )) = ax + b, using the hint above

T^-1(( 2b - a )*x + ( b+a )) = [?]x + 1/3 [( 2b - a ) + ( b+a )] or

T^-1(Qx +R) = (?)x + 1/3(Q+R) (by reducing the seconding coeffient to just be, think of a simple addition / subtraction you can do with:

Term 1 (2b - a) and Term 2 ( b+a ) to so that C*T1 + D*T2 = a and you have your answer for the [?]x component of the equation!

Like T1 - 2*T2 = ( 2b - a ) + -2( b+a ) = -3a,

so your first coffiecent is simply -1/3(T1 - 2T2) or

T^-1(ax + b) = -1/3(a - 2b)x + 1/3(a + b)

So check this out using T^-1(T(ax + b)) = ax + b to see I didn't do a typo somewhere:

T^-1(T(ax + b)) = T^-1(( 2b - a )*x + ( b+a )) =

-1/3([ 2b - a ] - 2[( b+a )])x + 1/3([( 2b - a )] + [( b+a )] ) =
-1/3(0-3a)x + 1/3(3b) =
ax + b

Solved!