# Thread: prove linear transformation is not surjective

1. ## 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

2. The endomorphism is:

$T:P(\mathbb{R})\rightarrow P(\mathbb{R})\,,\quad T[p(x)]=\int_0^x p(t)\;dt$

Suppose there exists $p(x)\in P(\mathbb{R})$ such that $T[p(x)]=1$ , use the Fundamental Theorem of Calculus and you will find a contradiction.

Fernando Revilla

3. 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 $F(x)\,,\,\,F(x)-F(0)$ is a polynomial with zero free coefficient...

Tonio

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

5. 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, $p(x)=0$ is not a polynomial, of course. It looks like an equation, though.

Tonio

6. Originally Posted by tonio
No, $p(x)=0$ 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 $P(\mathbb{R})$ fails to be a vector space. Could you clarify what you meant?

7. Originally Posted by roninpro
If we don't treat the zero function as a polynomial, then $P(\mathbb{R})$ 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...