Finding Basis of Nilpotent 
Consider the differential operator $T:P3(R)→P3(R)$ given by
$T(p(x))=2p′(x)−2p′′(x)+2p′′′(x)$
Find an ordered basis $F$ for $P3(R)$ such that $T$ acts like a shift
  operator with respect to $F$, i.e.
$M^{F}_{F} (P)$ =
  \begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}
Basis $F$ = {$ . . . $}
Enter your basis as polynomials separated by comma: e.g. $x2,2x+1,2$

So, I have a few similar questions like that and don't know how to solve them can someone help me with it.
 A: I assume $R$ is the field of real numbers, or at least a ring with unity, and $P^3(R)$ is the vectorspace of polynomials with coefficients in $R$ of degree atmost 3.
You can identify $P^3(R)$ with $R^4$ via
$$p(x) = ax^3 + bx^2 +cx +d \mapsto \begin{pmatrix} a \\ b \\c \\ d \end{pmatrix}$$
Under this identification the standard basis of $R^4$ corresponds to the polynomials $x^3, x^2, x ,1$. Problems like this can usually be treated by considering an isomorphism of this kind and reducing it to linear algebra on $R^n$ for some $n$. 
In the ordered basis $F = (x^3, x^2, x, 1)$ the map $p(x) \mapsto p'(x)$ can be identified with the matrix 
$$D = \begin{pmatrix}
0 & 0 & 0 & 0 \\
3 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 
\end{pmatrix}$$.
Your operator $T$ is equal to
$$T = 2 D - 2D^2 + 2 D^3.$$
Now you just need to find a basis in which $T$ is of the desired form.
Can you handle it from here on?
Edit
Plugging in the numbers we get the matrix
$$T = 
\begin{pmatrix}
0 & 0 & 0 & 0 \\
6 & 0 & 0 & 0 \\
-12 & 4 & 0 & 0 \\
12 & -4 & 2 & 0 
\end{pmatrix}$$
Let $F=(f_1,f_2,f_3,f_4)$ be a basis such that $T$ is as requested. Given the form of $T$ in this basis we know that under $T$ $f_1$ gets mapped to $f_2$, $f_2$ gets mapped to $f_3$, $f_3$ to $f_4$ and $f_4$ itself is in the kernel of $T$ (i.e $Tf_4 = 0$). This translates to a set of equations:


*

*$f_2 = Tf_1$ (first column)

*$f_3 = Tf_2 = T^2f_1$ (second column)

*$f_4 = Tf_3 = T^3f_1$ (third column)

*$Tf_4 = 0$ (fourth column)


We start by choosing $f_1 = (1,1,1,1)^T$, as there are no constraints on $f_1$ (except that it has to be linearly independent of the other basis elements of course). This yields
$$f_2 = Tf_1 = \begin{pmatrix} 0 \\ 6 \\ -8 \\ 10\end{pmatrix}, ~~f_3 = Tf_2 = \begin{pmatrix} 0 \\ 0 \\ 24 \\ -40\end{pmatrix} ~~ \text{and} ~~f_4 = Tf_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 48\end{pmatrix}$$
Now we have to check whether the fourth condition holds.
The kernel of $T$ is the linear span of the vector $(0, 0 , 0, 1)^T$, which means that $f_4$ has to be of the form $f_4 = (0, 0 , 0, \lambda)^T$ for some $\lambda \in R$. Clearly $f_4 \in \mathrm{ker}(T)$ and since $f_1$ up to $f_4$ are linearly independent we have found our basis. All that is left to do, is translating these vectors back to polynomials e.g. $f_1 = (1,1,1,1)^T$ corresponds to $x^3 + x^2 +x  + 1$. Note that you can start with any polynomial of degree 3 you like.
A: Interpreting the columns of the desired matrix form as the images of the basis vectors, you need to find a linearly-independent set of four polynomials $p_k$ such that $T(p_1)=p_2$, $T(p_2)=p_3$, $T(p_3)=p_4$ and $T(p_4)=0$.  
Clearly $p_4\in\ker(T)$, so you could start there. The kernel of $T$ consists of the constant polynomials, so you might as well choose $p_4=1$. Now you have to find a polynomial $p_3$ such that $T(p_3)=p_4$ and $p_3$ is not a constant polynomial. Observe that $T$ reduces the degree of a nonconstant polynomial by $1$, which means that $p_3=ax+b$ for some unknown coefficients $a$ and $b$. So, solve $T(ax+b)=1$ for $a$ and $b$ to get a $p_3$. Proceed in this same vein for the other two polynomials. At each step you’ll generate a polynomial of degree one greater than the previous one, so you’re assured that they’ll be linearly-independent. Note that the solutions to the equations might not be unique, so you’ll just need to pick whichever seems most convenient.  
Since $T$ reduces the degree of the polynomial by one, there’s an easier way to proceed: pick a cubic polynomial $p$ and repeatedly apply $T$ to it until you get to zero. This sequence of polynomials will be your basis. The simplest choice is $x^3$, of course.
