Kernel of Transpose of Linear Transformation is Annihilator of Image

Theorem
Let $G$ and $H$ be $n$-dimensional vector spaces over a field.

Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.

Let $u \in \map \LL {G, H}$.

Let $u^t$ be the transpose of $u$.

Then:
 * $\map \ker {u^t}$ is the annihilator of the image of $u$

where $\map \ker {u^t}$ denotes the kernel of $u^t$.

Proof
From the definitions of:
 * the transpose $u^t$
 * the annihilator $\paren {\map u G}^\circ$

it follows that:
 * $\map {u^t} y = 0 \iff y \in \paren {\map u G}^\circ$

Thus:
 * $\map \ker {u^t} = \paren {\map u G}^\circ$