Find the eigenvalues of this endomorphism of R[X]

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The discussion revolves around finding the eigenvalues of the endomorphism f defined on Rn[X] by f(P)(X) = ((aX+b)P)'. The participants clarify the implications of different values for a and b, noting that if a=0, the only eigenvalue corresponds to constant polynomials. When a is non-zero, the transformation leads to a differential equation that participants solve to find Q(X) and subsequently P(X). The conversation highlights the importance of ensuring that the exponent c/a - 1 is an integer for P to remain a polynomial. Ultimately, the key focus is on correctly interpreting the endomorphism and determining the conditions for valid eigenvalues.
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Homework Statement



f is an endomorphism of Rn[X]
f(P)(X)=((aX+b)P)'

eigenvalues of f?

Homework Equations



(a,b)<>(0,0)

The Attempt at a Solution



If a=0, then f(P)=bP', and only P=constant is solution

if a<>0, then I put Q=(ax+b)P, f(P)=cP is equivalent to (ax+b)Q'=Q (E)

I solved (E) and found Q(X)=(aX+b)^c but then if I say P(X)=(aX+b)^(c-1), I can't find c...
 
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Is f(P)=(aX+b)P' or [(aX+b)P]'? You wrote it both ways.
 


f(P)=[(aX+b)P]'
 


Sorry, I misread your initial post. I see what you did now. Your solution for Q(X) should be Q(X)=(aX+b)^{c/a} so P(X)=(aX+b)^{c/a-1}. Do you see now what values c can be?
 


Thanks a lot vela, I mistook when I solved (E)... Now I can see the values for c (s.t c/a-1 is integer and therefore P is a polynom)...
 
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