How to determine the base of $\ker\phi$ for polynomial function? Given is a base defined as $$B:=(x\mapsto1,x\mapsto x,x\mapsto x^2,x\mapsto x^3 ,x\mapsto x^4)$$ A set V defined as
$$V:= \{ f: \mathbb{R} \mapsto \mathbb{R}\ |\ \exists\ {a_0},...{a_4} \in \mathbb{R}\ : f(x)=\sum_{i=0}^{4}{a_ix^i} \ \forall \ x \in \mathbb{R}\}$$
a function $\phi$ defined as $$\phi(f)(x)=f''(x)-x \cdot f'(x) + f(x-1)$$
I determined the images of $\phi$ regarding the elements in the base B:
$\phi(1)=1$
$\phi(x)=-1$
$\phi(x^2)=-x^2-2x+3$
$\phi(x^3)=-2x^3-3x^2+9x-1$
$\phi(x^4)=-3x^4-4x^3+18x^2-4x+1$
I also calculated the following transformation matrix:
$M_B^B(\phi)=\begin{pmatrix} 1 & -1 & 3 & -1 & 1& \\ 0 & 0 & -2 & 9 & -4&\\ 0 & 0 & -1 & -3 & 18& \\ 0 & 0 & 0 & -2 & -4& \\ 0 & 0 & 0 & 0 & -3& \end{pmatrix}$
From this point on I don't know how to determine the base of $\ker\phi$.  I know the definition of $\ker\phi$ is $\ker\phi:=\{v \in V:\phi(v)=0\}.$ However I do not know how to apply this definition to my problem. 

You can ease the computation of the associated matrix by building the matrices of $f(x)\mapsto f''(x)$, $f\mapsto xf'(x)$ and $f(x)\mapsto f(x-1)$ so
$$
\begin{pmatrix}
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 6 & 0 \\
0 & 0 & 0 & 0 & 12 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
-
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0 & 4
\end{pmatrix}
+
\begin{pmatrix}
1 & -1 & 1 & -1 & 1 \\
0 & 1 & -2 & 3 & -4 \\
0 & 0 & 1 & -3 & 6 \\
0 & 0 & 0 & 1 & -4 \\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$
and find
\begin{pmatrix}
1 & -1 & 3 & -1 & 1\\
0 & 0 & -2 & 9 & -4\\
0 & 0 & -1 & -3 & 18\\
0 & 0 & 0 & -2 & -4\\
0 & 0 & 0 & 0 & -3\\
\end{pmatrix}
The RREF is
\begin{pmatrix}
1 & -1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
and a basis of the null space consists of the single vector
\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
The polynomial that has this vector as its coordinate vector is
$$
f(x)=1+x
$$
So a basis of the null space of $\phi$ consists of $1+x$.
