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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.

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    $\begingroup$ Hint: can you show that $\phi(x^2), \phi(x^3), \phi(x^4)$ are linearly independent? $\endgroup$ – Adam Higgins May 19 at 13:19
  • $\begingroup$ Can you elaborate a little on why it is necessary to prove, that $\phi(x^2), \phi(x^3), \phi(x^4)$ are linearly independent? $\endgroup$ – JulianGi May 19 at 13:24
  • $\begingroup$ Do you know how to solve a homogeneous linear system $Ax=0$ for some matrix $A$ using row transformations (Gaussian elimination)? $\endgroup$ – Christoph May 19 at 13:26
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    $\begingroup$ It's not necessary to prove it, but if you can see it, it will help you towards the answer. The rank-nullity theorem states that if $\phi : V \to W$ is a linear map of vector spaces $V,W$ with $\operatorname{dim}V$ finite, then the dimension of the kernel of $\phi$ and the dimension of the image of $\phi$ add up to the dimension of $V$. In this case, since we have three linearly independent vectors in the image, we see that the dimension of the image is at least $3$, and since the dimension of $V$ is $5$, it follows that the dimension of $\ker{\phi}$ is at most $2$. Can you finish? $\endgroup$ – Adam Higgins May 19 at 13:29
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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 $1+x$.

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  • $\begingroup$ Unfortunately I mixed up the definition of $\phi.$ I edited the original post. The correct definition is $\phi(f)(x)=f''(x)-x \cdot f'(x) + f(x+1).$ So my calculated matrix should be correct. $\endgroup$ – JulianGi May 19 at 13:58
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    $\begingroup$ @JulianGi No, it's wrong again, unless there's another mixup. $\endgroup$ – egreg May 19 at 14:20
  • $\begingroup$ I'm very sorry but there was indeed another mix up. The updated and now definitely correct definition of $\phi$ is $\phi(f)(x)=f''(x)-x \cdot f'(x) + f(x-1)$ $\endgroup$ – JulianGi May 19 at 14:23
  • $\begingroup$ Thank you very much for your patience I understood your solution now. $\endgroup$ – JulianGi May 19 at 14:38
  • $\begingroup$ @JulianGi It's your task to write a proper question and check it. Anyway, now the answer is final. $\endgroup$ – egreg May 19 at 14:38
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Let $v\in V$ and consider $\sum_{i=0}^4 v_i x_i$ its decomposition on the basis. $v$ is in $\ker \phi$ if and only if $$M_B^B(\phi)\begin{pmatrix}v_0\\v_1\\v_2\\v_3\\v_4\\\end{pmatrix}=0$$

This is a triangular system which is easily solved from end to start (begin with $v_4$). One finds $v_4=v_3=v_2=0$ and $v_0=v_1$. Hence $\ker \phi$ is spanned by the vector whose decomposition is $(1,1,0,0,0)$, that is $x\mapsto x+1$.

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  • $\begingroup$ Could you elaborate, how we can transform the definition of the kernel $\ker\phi:=\{v \in V:\phi(v)=0\}$ into $\ker\phi:=\{v \in V:M_B^B(\phi)\begin{pmatrix}v_0\\v_1\\v_2\\v_3\\v_4\\\end{pmatrix}=0\}$ $\endgroup$ – JulianGi May 19 at 13:43
  • $\begingroup$ @JulianGi See Theorem 0.23 in math.colorado.edu/~nita/MatrixRepresentations.pdf $\endgroup$ – Gabriel Romon May 19 at 13:47

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