Rank and Nullity of Transpose

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

Let $$\mathcal {L} \left({G, H}\right)$$ be the set of all linear transformations from $$G$$ to $$H$$.

Let $$u \in \mathcal {L} \left({G, H}\right)$$.

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

Then:
 * 1) $$u$$ and $$u^t$$ have the same rank and nullity;
 * 2) $$\mathrm{ker} \left({u^t}\right)$$ is the annihilator of the range of $$u$$;
 * 3) The range of $$u^t$$ is the annihilator of $$\mathrm{ker} \left({u}\right)$$.

Proof

 * From the definitions of the transpose $$u^t$$ and the annihilator $$\left({u \left({G}\right)}\right)^\circ$$, it follows that $$u^t \left({y'}\right) = 0 \iff y' = \left({u \left({G}\right)}\right)^\circ$$.

Thus $$\mathrm{ker} \left({u^t}\right) = \left({u \left({G}\right)}\right)^\circ$$.


 * Let $$x \in \mathrm{ker} \left({u}\right)$$.

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

Let $$\left \langle {x, t'} \right \rangle$$ be as defined in Evaluation Linear Transformation.

Then $$\forall y' \in H^*: \left \langle {x, u^t \left({y'}\right)} \right \rangle = \left \langle {u \left({x}\right), y'} \right \rangle = \left \langle {0, y'} \right \rangle = 0$$.

So $$u^t \left({H^*}\right) \subseteq \left({\ker u}\right)^\circ$$.

From Sum of Nullity and Rank of Linear Transformation and Results Concerning Annihilator of Vector Subspace:

$$ $$ $$ $$ $$

So it follows that $$u$$ and $$u^t$$ have the same rank and nullity, and that $$u^t \left({H^*}\right) = \left({\mathrm{ker} \left({u}\right)}\right)^\circ$$.