Transpose of the differentiation operator Please help me write down a step by step solution to the following problem  
Let $n$ be a positive integer and let $V$ be a finite dimensional vector space of all polynomial functions over the field of real numbers which have degree at most $n$ i.e. functions of the form $$f(x)=c_0+c_1x+\dots + c_nx^n$$  Let $D$ be the differentiation operator on $V$.Find a basis for the null space of the transpose operator $D^t$.
 A: We represent a polynomial of the form $f(x) = c_0 + c_1 x + \cdots + c_n x^n$ by the $n+1$ dimensional vector $(c_0,c_1,\cdots,c_n)^T$ and we first wish to find an $(n+1)\times(n+1)$-dimensional matrix $M$ to represent the differentiation operator $D$. Observing that $f'(x) = c_1 + 2 c_2 x + \cdots + n c_n x^{n-1}$, we can represent $f(x)$ by
$$D  ((c_0,c_1,\cdots,c_n)^T) = (c_1,2c_2,\cdots,n c_n,0)^T$$
And we can check that the matrix corresponding to this linear transformation has the form $$M = \begin{pmatrix}  0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 2 & \cdots & 0 \\ \vdots  & \vdots & \ddots & \vdots & \vdots \\ 0  & 0 &\cdots & 0 & n \\ 0 & 0& \cdots & 0  & 0 \end{pmatrix} $$ 
(i.e. we have $1,2,\cdots,n$ on the superdiagonal of the matrix). We see that the transpose of this matrix is $$M^T = \begin{pmatrix} 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 2 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & n & 0 \end{pmatrix} $$
and then we can see by inspection that any vector of the form $(0,\cdots,0,a)$ will be in the nullspace of $M^T$, so any vector of that form will give us the desired basis (we know that the nullspace of our $(n+1)\times(n+1)$-dimensional matrix will be 1-dimensional because our matrix has rank $n$).
In terms of polynomials of degree $n$, it is polynomials of the form $g(x) = c_n x^n$ which form the nullspace of the transpose of the differentiation operator.
