Rank and Nullity of Transpose

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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

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

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


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

$\blacksquare$


Sources

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