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1. The problem statement, all variables and given/known data

Let $n>1\in\, \mathbb{N}$. A map $A:\mathbb{R}_{n}[x]\to\mathbb{R}_{n}[x]$ is given with the rule $(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)$

a)Prove that this map is linear

b)Find some basis of the kernel

c)Find the dimension of the image

2. Relevant equations

$\mathbb{R}_{n}[x]$ is defined as the set of all polynomial with real coeficient that have the power less or equal to n

$kerA=\{x;Ax=0\}$ $imA=\{Ax,x\in\mathbb{R}_{n}[x]\}$ $p(x)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{2}x^2+a_{1}x^1+a_{0}x^0$

3. The attempt at a solution a)Proof that this map is linear $p,q\in A \\ (A(p+q))(x)=(x^n+1)(p+q)(1)+(p+q)^{'''}(x)=((x^n+1)(p)(1)+(p)^{'''}(x)+(x^n+1)(q)(1)+(q)^{'''}(x)=Ap(x)+Aq(x) \\ \text{ } \\ A(\theta p)(x)=\theta((x^n+1)(p)(1)+(p)^{'''}(x))=\theta(Ap)(x) $

c) I started doing this by writing out the possible polynomial $(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)\\ p(1)=a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\ldots+a_{2}+a_{1}+a_{0}\\ p^{'''}(x)=6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3} $

now I placed all of this together $ (Ap)(x)=(x^n+1)\displaystyle\sum_{i=0}^{n}a_{i}+6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3} $ Then I paired all of the same coeficients together and got $$ a_{n}(x^{n}+1+6\binom{n}{3}*x^{n-3}) \\ a_{n-1}(x^{n}+1+6\binom{n-1}{3}*x^{n-4}) \\ a_{n-2}(x^{n}+1+6\binom{n-2}{3}*x^{n-5}) \\ \vdots\\ a_{3}(x^n+7) \\ a_{2}(x^n+1) \\ a_{1}(x^n+1) \\ a_{0}(x^n+1) \\ $$ Here I noticed that the bottom 3 functions are linearly dependent, which means that If I want to find the basis or dimensions I should take $a_{0}\, and\, a_{1}$ out. Then I also noticed that all of the above (a[SUB]n[/SUB] to a[SUB]3[/SUB]) are also linearly dependent on a[SUB]2[/SUB] so I subtracted them and got $$ a_{n}(x^{n-3}) \\ a_{n-1}(x^{n-4}) \\ a_{n-2}(x^{n-5}) \\ \vdots\\ a_{3}(6) \\ a_{2}(x^n+1) \\ $$ as my basis for the image of A therefore the dimension(imA)=n-1

This is as far as I have gotten. I don't know if this is correct but I really don't know how to continue I would really appreciate if someone could show me how to solve b) and possibly check If what I did is even correct.

Thank you

  • 1
    If you already $\;\dim \text{Im}\,A=n-1\;$ , then it **must** be that $\;\dim\ker A=2\;$ , right? Remember the dimensions theorem... – 2017-01-15
  • 0
    Wouldn't that be the case only if $kerA$ and $imA$ have no common basis vector? However even If I know the dimension of the kernel I still don't know how to find the basis vectors – 2017-01-15
  • 1
    For example, $\;A(x-1)=A(x^2-1)=0\implies x-1,\,x^2-1\in\ker A\;$ ... – 2017-01-15
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    I understand that. However I'm still not sure how I would go about looking for the basis for this kind of example – 2017-01-15

1 Answers 1

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Your solution so far seems correct.

Using the dimension theorem - as commented - you will get the dimension of the kernel is 2.

However, you found a basis for the image, which was not asked and not for the kernel, which was asked.
Well, you can finish from here, just pick any 2 indepent vectors from the kernel.

a) For a fixed polynomial $q$ and number $x_0$, the functions $p\mapsto p(x_0)$ and also $p\mapsto p(x_0)\cdot q$ are linear. The differential $D:p\mapsto p'$ is also linear, hence so is $D\circ D\circ D=p\mapsto p'''$, and thus any linear combinations of them are linear.

b) Let's find the kernel. What does $Ap=0$ mean?
Observe that $\deg p'

So, the kernel consists of at most quadratic polynomials ($p'''=0$) that vanish at $1\ $ ($p(1)=0$). The polynomials which vanish at $1$ are exactly the multiples of $(x-1)$.

So, a basis for the kernel is: $( x-1, \ (x-1)\cdot x)$

c) Use the dimension theorem to obtain the $n-1$, in accordance with your basis $\big(6, x^n+1,\ x, x^2, x^3,\dots, x^{n-3} \big)$.