 Solution

The dual basis is $E^{\ast }=\{e_1^{\ast },\dots ,e_n^{\ast }\}$, where $e_i^{\ast }(e_j)={\delta }_{ij}$ and each coordinate linear functional extends linearly. Thus $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$. These functionals are a basis of $V^{\ast }$ because a linear functional is uniquely determined by its values on $E$.

For a linear subspace $U$ of dimension $k$, extend a basis $u_1,\dots ,u_k$ 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 $T^{\ast }:V^{\ast }\to V^{\ast }$ defined by $T^{\ast }f=f\circ T$. Composition of linear maps is linear, so $T^{\ast }f$ is a linear functional; and $T^{\ast }(af+bg)=a{T}^{\ast }f+b{T}^{\ast }g$, proving linearity in its input. If $T(e_j)={\sum }_i{a}_{ij}{e}_i$, thenThe coefficient of $e_j^{\ast }$ in $T^{\ast }{e}_i^{\ast }$ is consequently $a_{ij}$, so $\boxed{[T^{\ast }{]}_{E^{\ast }}=([T{]}_E{)}^{\mathsf{T}}}$. No inner product is involved.

A linear functional $f$ annihilates $\mathrm{Im}T$ exactly when $f(Tv)=0$ for all $v$, equivalently $T^{\ast }f=0$. Thus $(\mathrm{Im}T{)}^{\circ }=\ker T^{\ast }$. If $v\in \ker T$, then $(T^{\ast }f)(v)=f(Tv)=0$, giving $\mathrm{Im}{T}^{\ast }\subseteq (\ker T{)}^{\circ }$. The first equality and the dimension formula give $\dim \ker T^{\ast }=n-\dim \mathrm{Im}T$. By the rank-nullity theorem,The inclusion is therefore an equality: