Characteristic polynomial and polynomial vector space

Caims
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Homework Statement


Let V:= ℝ_{2}[t]
V \in f: v \mapsto f(v) \in V, \forall v \in V (f(v))(t) := v(2-t)
a) Check that f \in End(V)
b) Calculate the characteristic polynomial of f.

Homework Equations



The Attempt at a Solution


a) Is it sufficient to check that (f+g)(t)=f(t)+g(t) ?
b) Standard basis of polynomials is 1+t+t^2, so f(1)=(2-t) \\ f(t)=2t - t^2 \\ f(t^2)=2t^2-t^3

What should I do next? What's up with t^3 ? (The space is of polynomials of degree at most 2). How do I calculate this map into a matrix ?
 
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Caims said:

Homework Statement


Let V:= ℝ_{2}[t]
V \in f: v \mapsto f(v) \in V, \forall v \in V (f(v))(t) := v(2-t)
a) Check that f \in End(V)
b) Calculate the characteristic polynomial of f.

Homework Equations



The Attempt at a Solution


a) Is it sufficient to check that (f+g)(t)=f(t)+g(t) ?
b) Standard basis of polynomials is 1+t+t^2, so f(1)=(2-t) \\ f(t)=2t - t^2 \\ f(t^2)=2t^2-t^3

What should I do next? What's up with t^3 ? (The space is of polynomials of degree at most 2). How do I calculate this map into a matrix ?

Well, as you've observed, the given map doesn't take second degree polynomial into second degree polynomials, it maps them into the space of third degree polynomials. That wouldn't stop you from writing a matrix for it, but it won't be square. Not sure what to do with the rest of the question at all.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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