= Solution
{c}
For a <linear functional> $\lambda\in V'$, the definition of the <dual map> gives
$$
\lambda\in\ker T'
\iff \lambda(Tu)=0\text{ for every }u\in U
\iff \lambda|_{\operatorname{im}T}=0.
$$
Thus the <kernel of a linear map> is precisely the <annihilator of a vector subspace>
$$
\boxed{\ker T'=(\operatorname{im}T)^0.}
$$
Write $n=\dim U$, $m=\dim V$ and $r=\dim\operatorname{im}T$. The <rank-nullity theorem>, the <dimension> formula for an <annihilator of a vector subspace>, and $\dim V'=m$ give
$$
\dim\operatorname{im}T'=m-\dim\ker T'=m-(m-r)=r.
$$
Applying the same <dimension> formula in $U$ gives
$$
\dim(\ker T)^0=n-\dim\ker T=r.
$$
Finally, if $\alpha=T'\lambda$ and $u\in\ker T$, then $\alpha(u)=\lambda(Tu)=\lambda(0)=0$. Therefore $\operatorname{im}T'\subseteq(\ker T)^0$. These are finite-dimensional <vector subspaces> of equal <dimension>, so inclusion is equality:
$$
\boxed{\dim\operatorname{im}T'=\dim(\ker T)^0,\qquad
\operatorname{im}T'=(\ker T)^0.}
$$
This is the <dual image and kernel annihilator identity>.
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