Why is the derivitive function for polynomials a linear transformation that is onto but not one to one?

Question

Is it because the derivitives of different constants are zero such that if P(x)=0, the zero can be sent to more than one function when going the other direction? How do I know it's onto?

Answer 1

You are correct about the onto part. It is easy to show that it isn't one-to-one

Let f be the derivative function
Let p1 be (x+1)
Let p2 be (x+2)
then f(p1) = f(p2) = 1
So f is not one-to-one by definition

For the onto part, write out the general form of a polynomial.
cn * x^n +... + c0
You should write more terms than this to show the general pattern. It's tough to do subscripts here though so I didn't bother with typing it all.

Find the derivative of the general form.
Explain how all polynomials can be written in this form (the derivative form).

Answer 2

The derivative map f (which takes a polynomial p(x) to its first derivative p'(x)) is onto because every polynomial is the derivative of another polynomial (hence every element of the codomain of the derivative map is the image under f of some element in the domain: the definition of onto).

It is not one-to-one because p(x) = 3x and q(x) = 3x+1 are distinct elements of the domain which are both mapped to the polynomial r(x) = 3. Hence there are p and q in the domain such that p is not equal to q but f(p) = f(q).