Null Space of Differentiation Transformation What is the null space of differentiation transformation:
$\frac{\mathrm{d} }{\mathrm{d} x}:P_{n} \to P_{n}$
where $P_{n}$ is the space of all polynomials of degree $\leqslant n $ over the real numbers
What is the null space of the second derivative as a transformation of $P_{n}$ ? What is the null space of the kth derivative?
I am slightly at a loss here, as I realise that they are looking differentiation as a transformation, but not a simple algebraic one as that(e.g reflection over x axis or something like that, which can easily formulated into a matrix format). Can anyone :


*

*Give me some hints on how to approach this problem, specifically, express the differentiation as a matrix?

*Point me to some text/textbook which can help me build such concepts in a better way.


Further Edit:
$P_{n}$ is as rightly pointed out is a vector space made of linear combination of the basis set {$1,x,x^{2},...,x^{n}$}. So I reckon the y=d/dx p(x) = AX where,
$p'(x)=0$ for the null space of the transformation operator.
So one trivial solution is when p(x)=c 
Till now, this is what I could figure.
 A: As a linear operator on $P_n$, differentiation can be expressed in matrix terms, but doing so is completely unnecessary and obscures what’s really going on; you’re better off thinking in terms of the definition of the null space of a linear transformation.
The null space of an operator $T:V\to W$ is simply the set $\{v\in V:T(v)=0_W\}$ of vectors in $V$ that get sent to the $0$ vector of $W$ by $T$. For what polynomials $p(x)\in P_n$ is it true that $$\frac{d}{dx}p(x)$$ is the zero vector of $P_n$? (For starters, what is the zero vector of $P_n$?) This has a very simple answer that requires no fiddling with matrices.
Once you’ve handled the first derivative, the rest should be easy. To find the null space of $\frac{d^3}{dx^3}$, for instance, just ask yourself which $p(x)\in P_n$ have the property that $$\frac{d^3}{dx^3}p(x)$$ is the zero vector of $P_n$.
A: For general function spaces, the differentiation operator may be difficult to put into a matrix, but consider that $a_n x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ becomes $(n)a_nx^{n-1}+(n-1)a_nx^{n-2}+\dots+(1)a_1+0$. 
If you consider the initial polynomial to be the vector $[a_n, a_{n-1},\dots, a_1, a_0]^T$, there is actually a nice matrix representation for what just happened to it. Like Lim, I would encourage you to use small examples first.
Alternatively, don't forget the definition of nullspace of $D$ is all the elements $p(x)$ such that $D(p(x))=0$.
A: To write differentiation in $P_n$ as a matrix, first establish a basis for $P_n$. There are several (infinitely many, of course) but the most usual one is the basis $\left\{1,x,x^2,\dots,x^n\right\}$. Then the matrix of differentiation is an $(n+1)\times (n+1)$ matrix (notice that there are $n+1$ elements in the basis, so $P_n$ is actually $(n+1)$-dimensional), with entries $A_{ij}$ such that differentiating the $j^\mathrm{th}$ basis vector (i.e. $x^{j-1}$) produces $A_{ij}$ of the $i^\mathrm{th}$ basis vector (so for example since differentiating $x^2$ doesn't produce any $x^4$, we have $A_{42}=0$).
Edit: some of the other answerers definitely have a point when they say working out the null space of differentiation doesn't require matrices at all (thinking about linear maps in the abstract is sufficient and often simpler), but nevertheless if you want to see how to make one, this is what you have to do.
A: Hint:
A polynomial $a_0+a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1}$ living in $P_n$ can also be viewed as a vector $(a_0,a_1,\dots,a_{n-1})$ living in ${\bf R}^n$.
