Rank and Nullity of Transpose

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:
 * $(1): \quad u$ and $u^t$ have the same rank and nullity


 * $(2): \quad \map \ker {u^t}$ is the annihilator of the image of $u$


 * $(3): \quad$ The image of $u^t$ is the annihilator of $\map \ker u$.

Proof
From the definitions of the transpose $u^t$ and the annihilator $\paren {\map u G}^\circ$, it follows that:
 * $\map {u^t} {y'} = 0 \iff y' = \paren {\map u G}^\circ$

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

Let $x \in \map \ker u$.

Let $H^*$ be the algebraic dual of $H$.

Let $\innerprod x {t'}$ be the evaluation linear transformation.

Then:
 * $\forall y' \in H^*: \innerprod x {\map {u^t} {y'} } = \innerprod {\map u x} {y'} = \innerprod 0 {y'} = 0$

So:
 * $\map {u^t} {H^*} \subseteq \paren {\map \ker u}^\circ$

From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace:

So it follows that $u$ and $u^t$ have the same rank and nullity, and that:
 * $\map {u^t} {H^*} = \paren {\map \ker u}^\circ$