Let be the 6 dimensional vector space over consisting of the polynomials having degree at most 5. Let be the map given by , where and denote the first and second derivative of with respect to . (See this previous exercise.)

- Prove that is a linear transformation.
- We showed previously that the set is a basis of . Find bases for the image and kernel of with respect to this basis.

We begin with a lemma.

Lemma: Let be a commutative unital ring. Then given by is a module homomorphism. Proof: Let and let . Then . Thus , and so is a module homomorphism.

Thus it is clear that is a linear transformation.

Note the following:

Thus we see that the matrix of is

The reduced row echelon form of is the matrix

Since only the 1st, 2nd, 3rd, and 6th columns of are pivotal, we see that the set forms a basis for . Next, the solutions of have the form . Letting , we see that is a basis for .

## Comments

Are you sure of phi(x^5) ?

No; but it should be fixed now.

Thanks!

It’s not fixed yet. Still says phi(x^5) = 12x^5 instead of phi(x^5) = 2x^5

Alright- it should be fixed for real this time.