# 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:

- $(2): \quad$ $\ker \left({u^t}\right)$ is the annihilator of the image of $u$

- $(3): \quad$ The image of $u^t$ is the annihilator of $\ker \left({u}\right)$.

## 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:

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

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

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

Let $\left \langle {x, t'} \right \rangle$ be the 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:

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

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

\(\displaystyle \dim \left({u^t \left({H^*}\right)}\right)\) | \(=\) | \(\displaystyle n - \dim \left({\ker \left({u^t}\right)}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n - \dim \left({\left({u \left({G}\right)}\right)^\circ}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dim \left({u \left({G}\right)}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n - \dim \left({\ker \left({u}\right)}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dim \left({\left({\ker \left({u}\right)}\right)^\circ}\right)\) | $\quad$ | $\quad$ |

So it follows that $u$ and $u^t$ have the same rank and nullity, and that:

- $u^t \left({H^*}\right) = \left({\ker \left({u}\right)}\right)^\circ$

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 28$: Theorem $28.12$