Thread: prove linear transformation is not surjective

I know how to prove that it is one to one, but i don't know how to prove that it's not onto, can some1 help me ?

ps: P(R) is a polynomial

The endomorphism is:



Suppose there exists such that , use the Fundamental Theorem of Calculus and you will find a contradiction.


Fernando Revilla

Originally Posted by mystic

I know how to prove that it is one to one, but i don't know how to prove that it's not onto, can some1 help me ?

ps: P(R) is a polynomial

Hint: Check that for any polynomial is a polynomial with zero free coefficient...

Tonio

well isn't p(x) = 0 a polynomial ? so even we let T (p(x))= 1, p(x) = 0 still works right ?

Originally Posted by mystic

well isn't p(x) = 0 a polynomial ? so even we let T (p(x))= 1, p(x) = 0 still works right ?

No, is not a polynomial, of course. It looks like an equation, though.

Tonio

Originally Posted by tonio

No, is not a polynomial, of course. It looks like an equation, though.

Tonio

If we don't treat the zero function as a polynomial, then fails to be a vector space. Could you clarify what you meant?

Originally Posted by roninpro

If we don't treat the zero function as a polynomial, then fails to be a vector space. Could you clarify what you meant?

I meant the whole expression "p(x)=0" is not a pol. but an equation, of course.

And in this case it is the zero polynomial, not the zero function. Had he written "p(x):=0" things would

have been clearer (for me, at least).

Tonio

Ps. By the way, I think my first message shows easily and clearly why the given transf. isn't surjective, but

there hasn't been any discussion on it...