Exercise: Let $P_3$ denote a vector space of polynomial of degree 3 or lower (and real coefficients) and let the linear map $F:P_4\to P_4$ be given by $F(p(t))=((t^2+1)p(t))^n$. Find a basis of eigenvectors of $F$.

I don't really know how to start, since I don't think I really understand the question properly ;) First of all: why do they say "Let $P_3$ denote the vector space..." when they don't use that information anywhere else in the question formulation. Then they suddenly define a function which takes $P_4$ to $P_4$. I assume, by reading the definition of $P_3$, $P_4$ is a vector space of polynomial of maximum degree 4?

I wonder also if it is meant by $F(p(t))=((t^2+1)p(t))^n$ that $F(a_4t^4+a_3t^3+a_2t^2+a_1t+a_0)=((t^2+1)(a_4t^4+a_3t^3+a_2t^2+a_1t+a_0))^n$? So you just take a random 4'th degree polynomial and map it? When I understand everything that have been said so far, it's time to find a basis of eigenvectors! Wohoo. Problem is though, I don't really see how you do it when it's a vector space of polynomial. Do you proceed as usual? When I imagine doing this (notice imagine) it looks something like this:

$B=(\begin{bmatrix} a_0 & a_1 & a_2+a_0 & a_3+a_1 + a_4+a_2 & a_3 & a_4 \end{bmatrix} \begin{bmatrix} 1 \\ t \\ t^2 \\ t^3 \\ t^4 \\ t^5 \\ t^6 \end{bmatrix})^n$. Since $BX=\lambda X\iff BX-\lambda X=0\iff \lambda X-BX=0\iff (\lambda I-B)X=0$ and so I can "simply" take the determinant of $\lambda*I-B$, put it equal to $0$ (which feels kinda wrong in this case), find it's eigenvalues and lastly find it's eigenvectors, if i am allowed to take determinant of matrix B, that is (perhaps a stupid idea, but this was just my imaginations how i would do it otherwise). Well, I think this was my questions for now and as you've probably already mentioned when you've come this far in the text, yeah I am really confused... this makes my world turn upside down when you don't use numbers any longer...

Hopefully someone can help me understand this better. Thanks! :)

  • $\begingroup$ where do $n$ come from? $\endgroup$ – user251257 Aug 6 '15 at 17:04
  • $\begingroup$ It's stated exactly like it's stated in my exercise :/ $\endgroup$ – user231999 Aug 6 '15 at 17:22
  • 2
    $\begingroup$ in general $p^n$ is not linear. so something is missing $\endgroup$ – user251257 Aug 6 '15 at 17:25
  • $\begingroup$ Does it say anything before the exercise? Anything that might help here? $\endgroup$ – Omnomnomnom Aug 6 '15 at 17:55
  • $\begingroup$ what is the underlying field? $\endgroup$ – user251257 Aug 6 '15 at 21:04

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