= Solution
The <dual basis> is $E^*=\{e_1^*,\ldots,e_n^*\}$, where $e_i^*(e_j)=\delta_{ij}$ and each coordinate <linear functional> extends linearly. Thus $e_i^*(\sum_j a_je_j)=a_i$. These functionals are a <basis> of $V^*$ because a <linear functional> is uniquely determined by its values on $E$.
For a <linear subspace> $U$ of dimension $k$, extend a <basis> $u_1,\ldots,u_k$ to one of $V$. In its <dual basis>, the <annihilator of a vector subspace> $U^\circ$ is spanned by the last $n-k$ coordinate functionals. Hence $\boxed{\dim U+\dim U^\circ=\dim V}$.
The <transpose> in the source is the algebraic <dual map> $T^*:V^*\to V^*$ defined by $T^*f=f\circ T$. Composition of <linear maps> is linear, so $T^*f$ is a <linear functional>; and $T^*(af+bg)=aT^*f+bT^*g$, proving linearity in its input. If $T(e_j)=\sum_i a_{ij}e_i$, then
$$
(T^*e_i^*)(e_j)=e_i^*(Te_j)=a_{ij}.
$$
The coefficient of $e_j^*$ in $T^*e_i^*$ is consequently $a_{ij}$, so $\boxed{[T^*]_{E^*}=([T]_E)^{\mathsf T}}$. No <inner product> is involved.
A <linear functional> $f$ annihilates $\operatorname{Im}T$ exactly when $f(Tv)=0$ for all $v$, equivalently $T^*f=0$. Thus $(\operatorname{Im}T)^\circ=\ker T^*$. If $v\in\ker T$, then $(T^*f)(v)=f(Tv)=0$, giving $\operatorname{Im}T^*\subseteq(\ker T)^\circ$. The first equality and the dimension formula give $\dim\ker T^*=n-\dim\operatorname{Im}T$. By the <rank-nullity theorem>,
$$
\dim\operatorname{Im}T^*=\dim\operatorname{Im}T
=n-\dim\ker T=\dim(\ker T)^\circ.
$$
The inclusion is therefore an equality:
$$
\boxed{\operatorname{Im}T^*=(\ker T)^\circ.}
$$
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