Diagonalization of linear operators First of all, I´m sorry for my English, I´m Spanish so I hope you can all understand me.
Here is my problem. Let 
                             $T(p(x))=p(x+1)$
be a linear operator from the space of polynomials with real coefficients and degree less than or equal to $n$. I´m having trouble with the matrix of the operator, and without it, it is not possible to know if the operator is diagonalizable. $$$$
Thank you!
 A: If $\lambda$ is an eigenvalue of $T$ and $P$ is an eigenvector associated to $\lambda$ then
$$T(P(x))=P(x+1)=\lambda P(x)$$
If we assume that $\deg P\ge 1$ and $x_0$ is root of $P$ then $x_0+n,\; n\in\mathbb N$ is a root of $P$ which's impossible. We can see easily that a constant polynomial is an eigenvector associated to the eigenvalue $1$. Now it's clear that $T$ isn't diagonalizable.
A: (1) For the matrix, just take a basis of $\def\R{\mathbb R}\R[X]_{\le n}$, say $\mathcal B = \{1, \ldots, X^n\}$, compute its values under $T$, we have $$(X+1)^n = \sum_{k=0}^n \binom nk X^k$$
and hence 
$$ [T]_{\mathcal B} = \begin{pmatrix} \binom 00 & \binom 10 & \cdots & \binom n0\\
                                          0     & \binom 11 & \cdots & \binom n1 \\
                                         \vdots &           & \ddots & \vdots \\
                                              0 &           &    0   & \binom nn
\end{pmatrix} $$
(2) Another way to answer the original question is to look for eigenvalues and -vectors directly: Let $\lambda \in \R$, and $p(X) = \sum_{i=0}^n a_iX^i$ such that $Tp = \lambda p$. Looking at highest nonvanishing coefficient of $X^n$ on both sides, namely $a_{\deg p}$ and $\lambda a_{\deg p}$, we see that $\lambda = 1$ is the only possible eigenvalue. But if $Tp = p$, $p$ is periodic, and the only periodic polynomials are the constant ones, hence the dimension of $\ker (T - 1)$ is $1$. Hence $T$ is diagonizable only if $n = 0$.
A: If we consider the basis $\left\{1,x,x^2,\dots,x^n \right\}$, then if you apply the operator to $x^m$ you get
$$ T(x^m) = (1+x)^m = \sum_{k=0}^{m} {m\choose k} x^k,\quad k=0,1,\dots,m.$$
Now, based on the above formula, the first few rows in the representation matrix $A$ should be
$$ [1,0,0,\dots,0],$$
$$ [1,1 ,0,\dots, 0] $$
$$ [ 1,1,2,0,\dots, 0].$$
Now, you can find the general form of the matrix.
