<s>[INST] Determine the base of $\ker\phi$ and fibre of phi of a polynom function Good evening,
given is a base $B$ 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 $$\phi(f)(x)=f''(x)+x \cdot f'(x) - f(x+1)$$  (where $f'(x)$ is the first derivative and $f''(x)$ the second derivative)    
and an element $g, g \in V$ $$g(x)=3x^4+2x^3-x+1$$ 
The task is to find a base of the kernel of phi and to determine $\phi^{-1}(\{g\})$.
To determine a base of the $\ker\phi$ I need to find all functions that map to $0$. But before that I calculated the mapping rule (? not sure on the english word).
$$\phi(\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}) = \phi(\lambda_1\cdot\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix}+\lambda_2\cdot\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix}+\lambda_3\cdot\begin{pmatrix}0\\0\\1\\0\\0\end{pmatrix}+\lambda_4\cdot\begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix}+\lambda_5\cdot\begin{pmatrix}0\\0\\0\\0\\1\end{pmatrix})$$
The next step was to determine the coefficients $\lambda_1, ... ,\lambda_5$ which directly result of this matrix: $$\begin{pmatrix} 1 & 0 & 0 & 0 & 0&| & a \\ 0 & 1 & 0 & 0 & 0&| & b\\ 0 & 0 & 1 & 0 & 0&| & c \\ 0 & 0 & 0 & 1 & 0&|& d \\ 0 & 0 & 0 & 0 & 1&|& e \end{pmatrix}$$
$$\Rightarrow \lambda_1 = a, \lambda_2 = b, \lambda_3 = c, \lambda_4 = d, \lambda_5 = e$$
Knowing this we can continue finding the mapping rule:
$$\overset{\phi\ linear}= a\cdot\phi(\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix})+b\cdot\phi(\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix})+c\cdot\phi(\begin{pmatrix}0\\0\\1\\0\\0\end{pmatrix})+d\cdot\phi(\begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix})+e\cdot\phi(\begin{pmatrix}0\\0\\0\\0\\1\end{pmatrix})$$
$$=a\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}+b\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}+c\cdot\begin{pmatrix}1\\-2\\1\\0\\0\end{pmatrix}+d\cdot\begin{pmatrix}-1\\3\\-3\\2\\0\end{pmatrix}+e\cdot\begin{pmatrix}-1\\-4\\6\\-4\\3\end{pmatrix}$$  
$$=\begin{pmatrix}-a-b+c-d-e\\-2c+3d-4e\\c-3d+6e\\2d-4e\\3e\end{pmatrix}$$  
So far so good, now using the definition: $\ker\phi:=\{v \in V:\phi(v)=0\}.$
$$\begin{pmatrix}-a-b+c-d-e\\-2c+3d-4e\\c-3d+6e\\2d-4e\\3e\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\end{pmatrix}$$
$$\Rightarrow e=0,d=0,c=0,-a=b$$
Now I've got a problem choosing a base due to $-a=b$. As far as I can tell a base of $\ker\phi$ would have 2 elements in this case which aren't linear independent. I thought of: $$\{a\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}+b\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix} | a,b \in \mathbb{R}\}$$
For the other question I need to know the $\ker\phi$ since for $\forall v \in V, g\in \phi(g): $$\phi^{-1}({g})= v + \ker\phi$$
Anyway I calculated $v$ by solving this equation:
$$\begin{pmatrix}-a-b+c-d-e\\-2c+3d-4e\\c-3d+6e\\2d-4e\\3e\end{pmatrix}=\begin{pmatrix}1\\-1\\0\\2\\3\end{pmatrix}$$
$$\Rightarrow v =x^4+3x^3+3x^2-2$$
If you need further information simply ask in this thread. I hope I described the problem detailed enough.
 [/INST] Imo it'd be easier (or at least slightly less messier) as follows:
$$f(x)=\sum_{k=0}^4a_kx^k\implies $$
$$f(x)\in\ker\phi\iff f''(x)+xf'(x)=f(x+1)\iff$$
$$2a_2+(6a_3+a_1)x+(12a_4+2a_2)x^2+3a_3x^3+4a_4x^4=$$
$$=\sum_{k=0}^4a_k+\left(\sum_{k=1}^4ka_k\right)x+\left(a_2+3a_3+6a_4\right)x^2+\left(a_3+4a_4\right)x^3+a_4x^4$$
Comparing corresponding powers of $\;x\;$ we get:
$$4a_4=a_4\iff a_4=0\\3a_3=a_3\iff a_3=0\\2a_2=a_2\iff a_2=0\\a_1=a_1\\a_0=-a_1$$
We thus get that
$$f(x)\in\ker\phi\iff f(x)=a-ax\;,\;\;a\in\Bbb R\implies \dim\ker\phi=1\;,\;\;\ker\phi=\text{Span}\{1-x\}$$
</s>