Investigate how many endomorphisms $f \in L(\mathbb R[x]_{3})$ meet the conditions 
Investigate how many endomorphisms $f \in L(\mathbb R[x]_{3})$ meet the conditions: (i) $\ker f =  \operatorname{span}(1,x)$,  (ii) $f\circ f=f$,  (iii) $f(x^{2})=1-x+x^{2}$,  (iv) if $f(x^{3})=p$ then $p(1)=p'(1)=0$.Find a Jordan's matrix and basis for for each such endomorphism

I know that in (ii) $f \in L(\mathbb R[x]_{3})$ is a projection for a some subspace $U$ along the subspace $V$. But I think it is impossible to such a special observation is crucial in this task because it does not seem so complicated. At the same time, we did not do anything similar on the lectures, so I have no knowledge about it.
Can you help me?
 A: Conditions (i)-...-(v) have to be considered jointly. We are going to show that they determine a single  matrix.
We will consider the matrix $M$ describing the transformation with respect to the (canonical) basis $\{1,x,x^2,x^3\}.$
The columns of such a matrix associated with a linear operation are given by the images of the basis'elements. Let us show in a first step that : 
$$M=\begin{pmatrix}
0&0&1&d\\
0&0&-1&(c-2d)\\
0&0&1&(-2c+d)\\
0&0&0&c\end{pmatrix}.$$
Explanations :
Columns 1 and 2 : they are both zero because condition (i) says that $f(1)=0=0+0x+0x^2+0x^3$ and $f(x)=0=0+0x+0x^2+0x^3$.
Column 3 : (condition (iii)) coefficients of $1-1x+1x^2+0x^3$.
Column 4 : condition (iv) amounts to say that $1$ is a double root of $p$ ; thus the image can be written $p=(x-1)^2(cx+d)=d+(c-2d)x+(-2c+d)x^2+cx^3.$
Now, we must take into account the fact that $M^2=M$ (condition (ii)) with : 
$$M^2=\begin{pmatrix}
0&0&1&(d - 2c + cd)\\
0&0&-1&(2c - d + c^2 - 2cd))\\
0&0&1&(c + 1)(d-2c)\\
0&0&0&c^2\end{pmatrix}.$$
Identifiying $M^2$ and $M$ leads to a first condition $c^2=c$, thus necessarily $c=0$ or $c=1$. Let us consider each of these cases :


*

*if $c=0$, whatever the value of $d$, all other constraints are verified BUT we get in this way a fourth column proportional to the third one ;  thus we would have Rank(M)=1 ; this comes in contradiction with the somewhat hidden condition : $\dim \ker f = 2 \implies Rank(f)=4-2=2$. (I am indebted here to the solution by @Egreg : I hadn't seen at first this condition). Thus we must rule out this case. 

*if $c=1$, we deduce without difficulty that $d=2$ matches all conditions, giving the rank-2 unique solution :
$$M=\begin{pmatrix}
0&0&1&2\\
0&0&-1&-3\\
0&0&1&0\\
0&0&0&1\end{pmatrix}.$$
I think that the Jordan form will not be difficult. Btw, what are the eigenvalues of a projector ?
A: We know that $f(1)=0$, $f(x)=0$, $f(x^2)=1-x-x^2$; set $f(x^3)=a+bx+cx^2+dx^3$.
You might be confused by polynomial notation; here it might be convenient to set $v_1=1$, $v_2=x$, $v_3=x^2$ and $v_4=x^3$; then the above statements become more “linear algebra like”:
\begin{cases}
f(v_1)=0 \\
f(v_2)=0 \\
f(v_3)=v_1-v_2+v_3 \\
f(v_4)=av_1+bv_2+cv_3+dv_4
\end{cases}
Since $\{v_1,v_2,v_3,v_4\}$ is a basis for $\mathbb{R}_3[x]$, the above characterizes the linear map $f$, provided we determine $a,b,c,d$ so that conditions (i)–(v) are satisfied.
The matrix with respect to the basis $\{v_1,v_2,v_3,v_4\}$ is
$$
A=\begin{bmatrix}
0 & 0 & 1 & a \\
0 & 0 & -1 & b \\
0 & 0 & 1 & c \\
0 & 0 & 0 & d
\end{bmatrix}
$$
We can compute $A^2$:
$$
A^2=
\begin{bmatrix}
0 & 0 & 1 & ad+c \\
0 & 0 & -1 & bd-c \\
0 & 0 & 1 & cd+c \\
0 & 0 & 0 & d^2
\end{bmatrix}
$$
Since we need $A=A^2$, we deduce
\begin{cases}
ad+c=a \\
bd-c=b \\
cd+c=c \\
d^2=d
\end{cases}
that immediately splits into $d=0$ or $d=1$.
Case $d=0$
We get $a=c$ and $b=-c$. Therefore $f(v_4)=c(v_1-v_2+v_3)$. In terms of polynomials, we have $f(v_4)=p$, where $p(x)=c(1-x+x^2)$. We have $p(1)=c$ and condition (v) forces $c=0$. This cannot be the case, because the rank would be $1$, contrary to the requirement that $\dim\ker f=2$.
Case $d=1$
We get $c=0$. Thus $f(v_4)=av_1+bv_2+v_4$ and so $p(x)=a+bx+x^3$. The condition $p(1)=0$ yields $a+b+1=0$; the condition $p'(1)=0$ yields $b+3=0$, so $b=-3$ and $a=2$.
Final discussion
The matrix of $f$ is
$$
A=\begin{bmatrix}
0 & 0 & 1 & 2 \\
0 & 0 & -1 & -3 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
The matrix has a double eigenvalue $0$ and a double eigenvalue $1$. Since also the eigenspace relative to $1$ has dimension $2$, the matrix is diagonalizable.
