# 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 $u^\intercal$ be the transpose of $u$.

Then:

$u$ and $u^\intercal$ have the same rank and nullity

## Proof

 $\ds \map \dim {\map {u^\intercal} {H^*} }$ $=$ $\ds n - \map \dim {\map \ker {u^\intercal} }$ $\ds$ $=$ $\ds n - \map \dim {\paren {\map u G}^\circ}$ $\ds$ $=$ $\ds \map \dim {\map u G}$ $\ds$ $=$ $\ds n - \map \dim {\map \ker u}$ $\ds$ $=$ $\ds \map \dim {\paren {\map \ker u}^\circ}$

Hence it follows that $u$ and $u^\intercal$ have the same rank and nullity.

$\blacksquare$