1. Derivative linear transformation

Let $\displaystyle P_3$ be a space of all polynomials (with real coefficients) of degree at most 3. Let $\displaystyle D : P_3 -> P_3$ be the linear transformation given by taking the derivative of a polynomial.

That is $\displaystyle D(a + bx + cx^2 + dx^3) = b + 2cx + 3dx^2$

let $\displaystyle \beta$ be the standard basis $\displaystyle (1,x,x^2,x^3)$ of $\displaystyle P_3$.

Find the matrix $\displaystyle M_D$ of $\displaystyle D$ with respect to the standard basis.

I'm sure that this isn't too difficult, but I can't get my head around how to start it. I decided to find D by letting it be the transformation to get from the standard basis to the derivative.

so $\displaystyle M_D M_{\beta}=[b + 2cx + 3dx^2]$

So $\displaystyle M_D = [b$ $\displaystyle 2c$ $\displaystyle 3d$ $\displaystyle 0]$. But I have a feeling that it's not at all that simple.
Any input is greatly appreciated.

2. You just gotta check the action of the basis elements under the map.

For instance the first basis element 1 goes to 0 the derivative of 1 is
$\displaystyle D(1)= 0 = 0 + 0x + 0x^2 + 0x^3$
$\displaystyle D(x)= 1 = 1 + x + 0x^2 + 0x^3$
$\displaystyle D(x^2)= 2x = 0 + 2x + 0x^2 + 0x^3$
$\displaystyle D(x^3)= 3x^2 = 0 + 0x + 3x^2 + 0x^3$

First column all zeros
second column has a 1 in the first spot then all 0s
Third column has a 2 in the second spot 0s elsewhere
4th has a 3 in the 3rd spot and 0s elsewhere

that is 1,2,3 on the first superdiagonal.

Check its action on an arbitrary degree 3 polynomial
$\displaystyle [a,b,c,d]=a+bx+cx^2+dx^3$
Looks good to me.

3. Shouldn't I be able to multiply the matrix you've given by matrix $\displaystyle [1;x;x^2;x^3]$ to get the derivative?

4. yeah, I am not sure you are understanding this basis.

The basis itself is $\displaystyle \{1,x,x^2,x^3\}$. So when you are looking at the action of the derivative matrix on a given polynomial, you put the column vector consisting of the coefficients on the polynomial on the right side of the matrix and multiply. So for that one you supplied, I am assuming you are wanting the derivative of $\displaystyle 1+x+x^2+x^3$, you would put the column vector consisting of all ones.

What you get out will be [1,2,3,0] but that is telling you the derivative is
$\displaystyle 1*1+2*x+3*x^2 + 0*x^3$ which is consistent with the derivative. Got it?

Like if these were just regular vectors in $\displaystyle \mathbb{R}^3$ you wouldn't put $\displaystyle \{\hat{i},\hat{j},\hat{k}\}$ in there to look at the transformation, you would put the coefficients. That is the whole point of representing a linear transformation in terms of a matrix.

5. Ok thanks, with that second post it makes complete sense.

In my first post I wrote out the question word for word.
The next part asks for the rank and nullity of D.
I can see the rank of $\displaystyle M_D$ is going to be 3 and consequently the nullity will be 1.
Will it be the same for D, I'm having a little trouble understanding the difference between D and $\displaystyle M_D$.

Is D the linear transformation itself, and $\displaystyle M_D$ the matrix that represents this transformation?
If so, then the rank and nullity for D should be the same as $\displaystyle M_D$ right?
If so I'm going to be ever so happy.

Cheers.

6. Oh yeah definitely. The two are pretty much defined to be exactly one another. It is like the rank of the linear transformation is the dimension of the image as a vector space and the nullity is the dimension of the kernel as a vector space. But that is exactly what the matrix shows you. The kernel is all the stuff that goes to 0, ie the polynomials of degree 0 (the second, third and fourth entries in the vector that is getting acted on must be 0 or else you result is not the 0 vector) which is of dimension one (it is just the real numbers), so that is your null space. The rank is the stuff spanned by this matrix which clearly only goes up to polynomials with degree 2, 1 or 0 so it has dimension 3.

Spot on.