Geometric intuition of adjoint For a linear operator it holds that $\ker (T^\ast ) = (\operatorname{ran} (T))^\perp$. The star denote the adjoint of $T$ and $\perp$ the orthogonal complement. Is there a geometric intuition for the meaning of $\ker (T^\ast ) = (\operatorname{ran} (T))^\perp$? 
 A: There is a related question on an intuitive view on adjoints at mathoverflow. Although not directly related to $\ker T^*=(\operatorname{ran}T)^\perp$, it may help your geometric intuition:
Considering the graph of $T:\mathcal{H}\to\mathcal{H}$, $\mathcal{G}(T):=\{(x,Tx)\mid x\in\mathcal{H}\}$ we can naturally equip the product space $\mathcal{H}\times\mathcal{H}$ with an inner product by setting $\langle(x,y),(x',y')\rangle:=\langle x,x'\rangle+\langle y,y'\rangle$ and therefore look at the orthogonal complement of $\mathcal{G}(T)\subset\mathcal{H}\times\mathcal{H}$ in the ambient product space $(y,z)\in\mathcal{G}(T)^\perp$:
$$0=\langle(x,Tx),(y,z)\rangle=\langle x,y\rangle+\langle Tx,z\rangle=\langle x,y\rangle+\langle x,T^*z\rangle\qquad\Rightarrow\qquad y=-T^*z$$
Thus, the adjoint $T^*$ is the operator such that $\mathcal{G}(T^*)=-(\mathcal{G}(T))^\perp$, i.e. the operator which (up to an sign) has the graph orthogonal to $T$.
A: For the sake of simplicity assume we have an inner product so the dual space of each vector space is naturally isomorphic to the vector space.
Let $T: V \to W$ be a homomorphism between vector spaces and $T' : W' \to V'$ be the adjoint. Now the image of adjoint $\mathrm{Im}\;T'$ is all non-negligible elements of the domain $\mathrm{Dom}\;T$ of $T$.
If none-of the elements of $\mathrm{dom}\;T$. are negligible, then 
$$ \mathrm{Im}\;T' =  \mathrm{Dom}\;T
$$
In fact $T'$ is surjective whenever $T$ is injective and vice versa. In other words, the image of $T'$ is $V$ as seen from $W$ through the lens of $T$.
