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My first thought on solving this is defining $f=a_2x^2+a_1x+a_0$

Now using $T(f)=f''+f'$

$T(a_2x^2+a_1x+a_0)=2a_2x+a_1+2a_2=\lambda a_2x^2+ \lambda a_1x+ \lambda a_0$

After that I get the next system of equations: $$\lambda a_2=0$$ $$\lambda a_1=2a_2$$ $$\lambda a_0=a_1+2a_2$$ Which gives me $\lambda =0$, after this I substitute $\lambda$ in $f=\lambda a_2x^2+ \lambda a_1x+ \lambda a_0=0$. Can I say my eigenvector and eigenvalue $=0$ ?

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The eigenvalue is indeed zero, as can plainly be seen by the fact that $T$ reduces any polynomial by at least one degree. However there are more eigenvectors (i.e. 0 has multiplicity). Specifically you'd need to solve $Tf=0$, for which there are plenty of solutions.

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