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:
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.12$