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.

• Hint: can you show that $\phi(x^2), \phi(x^3), \phi(x^4)$ are linearly independent? – Adam Higgins May 19 at 13:19
• 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? – JulianGi May 19 at 13:24
• Do you know how to solve a homogeneous linear system $Ax=0$ for some matrix $A$ using row transformations (Gaussian elimination)? – Christoph May 19 at 13:26
• 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? – Adam Higgins May 19 at 13:29

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 $$1+x$$.

• 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. – JulianGi May 19 at 13:58
• @JulianGi No, it's wrong again, unless there's another mixup. – egreg May 19 at 14:20
• I'm very sorry but there was indeed another mix up. The updated and now definitely correct definition of $\phi$ is $\phi(f)(x)=f''(x)-x \cdot f'(x) + f(x-1)$ – JulianGi May 19 at 14:23
• Thank you very much for your patience I understood your solution now. – JulianGi May 19 at 14:38
• @JulianGi It's your task to write a proper question and check it. Anyway, now the answer is final. – egreg May 19 at 14:38

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 $$x\mapsto x+1$$.

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