# Let $V= \mathbb{R}_3[x]$ be the vector space of polynomials

1. Oct 20, 2011

### Ted123

Let $V= \mathbb{R}_3[x]$ be the vector space of polynomials with real coefficients with degree at most 3 and let $D:V\to V$ be the linear operator of taking derivatives, $D(f)=f'$.

I'm trying to check the Rank-nullity theorem for this example but it doesn't seem to hold:

Since $D$ is not injective (e.g. images of $x^2$ and $x^2+1$ are both $2x$), $\text{Ker}(D)\neq \{ \bf{0} \}$. (The kernel is actually all the constant polynomials).

Since $D$ is surjective (as every polynomial is the derivative of some other polynomial), $\text{Im}(D)=V=\mathbb{R}_3[x]$.

Now the Rank-nullity formula is:

$\text{dim}(\text{Ker}(D)) + \text{dim}(\text{Im}(D)) = \text{dim}(V)$.

But $\text{dim}(\text{Im}(D)) = \text{dim}(\mathbb{R}_3[x]) = 4 = \text{dim}(V)$

so if the Rank-nullity formula holds this would imply $\text{dim}(\text{Ker}(D))=0$ but it doesn't as $D$ is not injective, so what's going on?

EDIT: D is not surjective! How do I find a basis for the Kernel and Image?

Last edited: Oct 20, 2011
2. Oct 20, 2011

### Staff: Mentor

Re: Rank-nullity

Your mistake is here. Im(D) != R3. The image of a polynomial of degree 3 is a polynomial of degree 2, so Im(D) does not contain 3rd degree polynomials, hence D is not onto (surjective).

BTW, the notation used in this problem seems nonstandard to me. The space of real polynomials of degree less than n is often called Pn. So your space would normally be called P4. At least, that's the notation I've seen most often.

3. Oct 21, 2011

### Ted123

Re: Rank-nullity

Yeah I realised that restricting the degrees of the polynomials made D not surjective.

So with this notation does $\text{Im}(D) = \mathbb{R}_2[x]$ (i.e. Polynomials in x with real coefficients with degree at most 2)?

We know $\text{Ker}(D)=\{ \text{constant polynomials} \}$ so to describe all the bases of $\text{Ker}(D)$ can I just say the set of singletons $\{k\}$ where $k\in\mathbb{R} \setminus \{0\}$ ? So $\text{Ker}(D)$ has dimension 1.

And $\text{Im}(D) = \mathbb{R}_2[x]$ so a basis for the image is $\{1,x,x^2 \}$ with dimension 3. Then 1+3=4=dim(V) so the Rank-nullity formula is satisfied.

4. Oct 21, 2011

### Staff: Mentor

Re: Rank-nullity

Yes.
Yes, Ker(D) has dimension 1. You don't need to (and shouldn't) exclude the 0 polynomial.