# Can someone check if I did this right?

1. Feb 6, 2005

### eckiller

Q. Consider the case with V being the kth order polynomials with real coefficients. Let the derivative mapping D be the transformation which assigns to each polynomial function its derivative. Show that D maps V into V. What is the rank, nullity, nullspace, and range of D?
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This is what I did:

Let p = a_0 + a_1 x + a_2 x^2 +...+ a_k x^k in V.

D(p) = a_1 + 2 a_2 x + ... + k a_k x^(k-1).

So D(p) in V since it is a polynomial of at most k.

Now the thing with the rank and nullity, is there suppose to be a rigorous way to show these? The only way I know how to find them is by "eyeballing" the space.

I note that only constants and the zero polynomial have zero derivatives, hence N(T) = { a_0 | a_0 in Reals }.

And the range R(T) = {p(x) | p(x) = a_1 + 2 a_2 x + k a_k x^(k-1) }

Rank(T) = k
Nullity(T) = 1
Dim(V) = k + 1

Thanks for checking my work.

2. Feb 6, 2005

### AKG

Of at most what? Degree? That would be wrong, since they are of degree at most k-1. You're right, D(p) is in V, and I suppose you've shown it, but it's not technically correct to say, "it is a polynomial of at most (degree) k", rather, you should say that it is a polynomial of degree no greater than k, thus it is in V.
You want to find the polynomials whose derivative is zero. Can't you rigourously prove that only constant polynomials have derivative zero? Pick an arbitrary polynomial in V, take it's derivative, recognize that it is a sum of linearly independent vectors {1, x, x², x³, ... x^(k-1)} and so for it to be zero, the coefficients must be zero. Conclude that if all those coefficients must be zero, then the original polynomial must be constant.
Again, you can easily prove this. Take an arbitrary element of V and show that T maps it into the space that you've stated above. Also show that any element in that space (notice that it is the space of polynomials of degree at most k-1) is the image of some element of V. This shows that R(T) is a subset of the k-1 polynomials, and that the k-1 polynomials is a subset of R(T), showing that the two sets are equal as you've asserted.

You've done everything right, just make sure you take the time to prove it.

3. Feb 6, 2005

### eckiller

Great...thanks very much. That helped clear some things up.