= Solution
Because the <linear functionals> $\ell_0,\ell_1$ agree with endpoint <point evaluations> on <continuous functions>,
$$
\ell_i(e_j)=\delta_{ij}\qquad(i,j\in\{0,1\}).
$$
The endpoint values also make $e_0,e_1$ <linearly independent>. By its definition, the <image of a linear map> $T$ lies in $E=\langle e_0,e_1\rangle$. Conversely $Te_j=e_j$, so $T$ acts as the <identity operator> on $E$. Thus
$$
\boxed{\operatorname{Im}T=\langle e_0,e_1\rangle,\qquad\operatorname{rank}T=2.}
$$
For every $f$, $Tf\in E$, whence $T^2f=Tf$. It is a <linear projection>, and an annihilating <polynomial> is
$$
\boxed{p(z)=z(z-1),\qquad p(T)=T^2-T=0.}
$$
Back to article page
