Consider the vector space of polynomials of degree 2 or lower with real coefficient $\mathbb{P}_2(\mathbb{R})$, operator $$T(f(x)) = f''(x)+f'(x)+f(0)x^2$$ and vector $g(x) = x$.
Determine the characteristic polynomial of $T$ without constructing a representation matrix of $T$.
My try: I determined the T-cyclic subspace $W$ generated by $g(x)$, which is $span\{x, 1, x^2\}$. I know a theorem that says the characteristic polynomial of $T_w$ is $$(-1)^k(a_0+a^1t+...+a_{k-1}t^{k-1}+t^k)$$. But how can find the coefficients $a_1,a_2,...$?
Thank you!
We compute $T^3$ by beginning with the square. Let $f \in \mathbb R_2[X]$. We note that $f''' = 0$. We have :
$$T^2(f) = 2 f(0) + f'' + 2 f(0)X + (f''(0) + f'(0))X^2 $$ and $$ T^3(f) = 2(f''(0) + f'(0)) + 2f(0)+2 (f''(0) + f'(0)) X + (2f(0) +f''(0))X^2$$ Using $$f''(0) = f'' \quad \quad f = f(0) + f'(0) X + \frac{1}{2}f''(0) X^2 \quad \quad f' = f'(0) + f''(0) X$$ we obtain $$ T^3(f) = 2f'' + 2f' + 2f + f(0) X^2$$
So $T^3 = 2T + 2\mathrm{id}$. It is not hard to see that $ X^3 - 2X - 2$ is in fact the minimum polynomial of $T$. It is then its characteristic polynomial ($k=3$).